% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
%                                                                         %
% The Project Gutenberg EBook of Euclid's Book on Divisions of Figures, by 
% Raymond Clare Archibald                                                 %
%                                                                         %
% This eBook is for the use of anyone anywhere at no cost and with        %
% almost no restrictions whatsoever.  You may copy it, give it away or    %
% re-use it under the terms of the Project Gutenberg License included     %
% with this eBook or online at www.gutenberg.org                          %
%                                                                         %
%                                                                         %
% Title: Euclid's Book on Divisions of Figures                            %
%                                                                         %
% Author: Raymond Clare Archibald                                         %
%                                                                         %
% Release Date: January 21, 2012 [EBook #38640]                           %
%                                                                         %
% Language: English                                                       %
%                                                                         %
% Character set encoding: ISO-8859-1                                      %
%                                                                         %
% *** START OF THIS PROJECT GUTENBERG EBOOK EUCLID'S BOOK ON DIVISIONS ***%
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The Project Gutenberg EBook of Euclid's Book on Divisions of Figures, by 
Raymond Clare Archibald

This eBook is for the use of anyone anywhere at no cost and with
almost no restrictions whatsoever.  You may copy it, give it away or
re-use it under the terms of the Project Gutenberg License included
with this eBook or online at www.gutenberg.org


Title: Euclid's Book on Divisions of Figures

Author: Raymond Clare Archibald

Release Date: January 21, 2012 [EBook #38640]

Language: English

Character set encoding: ISO-8859-1

*** START OF THIS PROJECT GUTENBERG EBOOK EUCLID'S BOOK ON DIVISIONS ***
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%%%% Credits and transcriber's note %%%%
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Produced by Joshua Hutchinson, Ralph Carmichael and the
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\Large
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\textbf{EUCLID'S BOOK}
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\textbf{ON DIVISIONS OF FIGURES}
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\large \textsc{cambridge university press}

\textsc{c.~f.~clay}, \normalsize\textsc{Manager}

$\textgoth{London:}$ FETTER LANE, E.C.

$\textgoth{Edinburgh:}$ 100 PRINCES STREET

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\textgoth{Bombay, Calcutta\; and\; Madras:} MACMILLAN AND CO., \textsc{Ltd.}

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\Huge
\textbf{EUCLID'S BOOK}

\textbf{ON DIVISIONS OF FIGURES}

\medskip

\large
(\ItalicGreekSnippet{per`i diair'esewn bibl'ion})

\bigskip

\Large
WITH A RESTORATION BASED ON

WOEPCKE'S TEXT

\bigskip
\normalsize
AND \ ON \ THE
\bigskip

\Large
\textit{PRACTICA \ GEOMETRIAE}
\medskip

OF LEONARDO PISANO\\[27mm]
\normalsize
BY

\Large
RAYMOND CLARE ARCHIBALD, \textsc{Ph.D.}
\medskip

\small
ASSISTANT PROFESSOR OF MATHEMATICS IN BROWN

UNIVERSITY, PROVIDENCE, RHODE ISLAND

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\Large
Cambridge:

at the University Press

1915.

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$\textgoth{Cambridge:}$
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\textsc{printed by john clay, m.~a.}
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\textsc{at the university press}
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\pdfbookmark[0]{Dedication}{Dedication}
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\Large
TO
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MY OLD TEACHER AND FRIEND
\bigskip
\huge

ALFRED DEANE SMITH

\bigskip

\large
PROFESSOR OF GREEK AND LATIN
\bigskip

AT MOUNT ALLISON UNIVERSITY
\bigskip

FOR FORTY-FOUR YEARS
\bigskip

SCHOLAR OF GREAT ATTAINMENTS
\bigskip

THE WONDER OF ALL WHO KNOW HIM
\bigskip

THESE PAGES ARE AFFECTIONATELY DEDICATED

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\fancyhead[C]{\Heading{INTRODUCTORY}}
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\LargeCenteredNote{INTRODUCTORY}
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\pdfbookmark[0]{Introductory}{Introductory}

Euclid, famed founder of the Alexandrian School of
Mathematics, was the author of not less than nine
works. Approximately complete texts, all carefully edited,
of four of these,
(1)~the \textit{Elements},
(2)~the \textit{Data},
(3)~the \textit{Optics},
(4)~the \textit{Phenomena}, are now our possession.
In the case of
(5)~the \textit{Pseudaria},
(6)~the \textit{Surface-Loci},
(7)~the \textit{Conics}, our
fragmentary knowledge, derived wholly from Greek sources,
makes conjecture as to their content of the vaguest nature.
On~(8) the \textit{Porisms}, Pappus gives extended comment.
As to~(9), the book \textit{On Divisions} (\textit{of figures}), Proclus alone among
Greeks makes explanatory reference.  But in an Arabian
MS., translated by Woepcke into French over sixty years ago,
we have not only the enunciations of all of the propositions
but also the proofs of four of them.

Whilst elaborate restorations of the \textit{Porisms} by Simson
and Chasles have been published, no previous attempt has
been made (the pamphlet of Ofterdinger is not forgotten) to
restore the proofs of the book \textit{On Divisions} (\textit{of figures}). And,
except for a short sketch in Heath's monumental edition of
Euclid's \textit{Elements}, nothing but passing mention of Euclid's
book \textit{On Divisions} has appeared in English.

In this little volume I have attempted:

\renewcommand{\labelenumi}{(\arabic{enumi})}
\begin{enumerate}
\item to give, with necessary commentary, a restoration of Euclid's
work based on the Woepcke text and on a thirteenth century geometry
of Leonardo Pisano.

\item to take due account of the various questions which arise in
connection with (\textit{a}) certain MSS. of ``Muhammed Bagdedinus,''
(\textit{b}) the Dee-Commandinus book on divisions of figures.

\item to indicate the writers prior to 1500 who have dealt with
propositions of Euclid's work.

% -----File: 011.png---Folio vii------------------------------------------------------
\item to make a selection from the very extensive bibliography of the
subject during the past 400 years.
\end{enumerate}

In the historical survey the MSS. of ``Muhammed Bagdedinus''
play an important rôle, and many recent historians,
for example Heiberg, Cantor, Hankel, Loria, Suter, and
Steinschneider, have contributed to the discussion. As it is
necessary for me to correct errors, major and minor, of all of
these writers, considerable detail has to be given in the first
part of the volume; the brief second part treats of writers on
divisions before 1500; the third part contains the restoration
proper, with its thirty-six propositions. The Appendix deals
with literature since 1500.

A score of the propositions are more or less familiar as
isolated problems of modern English texts, and are also to be
found in many recent English, German and French books
and periodicals. But any approximately accurate restoration
of the work as a whole, in Euclidean manner, can hardly fail
of appeal to anyone interested in elementary geometry or in
Greek mathematics of twenty-two centuries ago.

In the spelling of Arabian names, I have followed Suter.

It is a pleasure to have to acknowledge indebtedness to
the two foremost living authorities on Greek Mathematics.
I refer to Professor J.~L. Heiberg of the University of
Copenhagen and to Sir Thomas~L.\ Heath of London.
Professor Heiberg most kindly sent me the proof pages of
the forthcoming concluding volume of Euclid's \textit{Opera Omnia},
which contained the references to Euclid's book \textit{On Divisions
of Figures}. To Sir Thomas my debt is great. On nearly
every page that follows there is evidence of the influence of
his publications; moreover, he has read this little book in
proof and set me right at several points, more especially in
connection with discussions in Note~113 and Paragraph~50.

\hfill R.~C.~A. \qquad\qquad

\quad\textsc{Brown University},

\quad\phantom{\textsc{Brown}}\textit{June}, 1915.
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\thispagestyle{empty}
\LargeCenteredNote{CONTENTS}
\pdfbookmark[0]{Contents}{Table of Contents}
\small
\noindent
\begin{tabular}{clr}
&&\textsc{page}\\
 &  \textsc{Introductory} & vi \\
\multicolumn{3}{c}{I} \\
\multicolumn{2}{l}{\textsc{paragraph}} & \\
\multicolumn{2}{l}{\textsc{ numbers}} & \\

1 & \textsc{Proclus, and Euclid's Book} \textit{On Divisions of Figures}  &  \pageref{par1} \\
2--6 & \textit{De Divisionibus}
      \textsc{by ``Muhammed Bagdedinus'' and} & \\
    &  \textsc{the Dee MS.} &     \pageref{par2} \\
7--9 & \textsc{The Woepcke-Euclid} MS.  &    \pageref{par7} \\
10--13 & \textit{Practica Geometriae} \textsc{of Leonardo Pisano (Fibonaci)} &     \pageref{par10} \\
14--17 & \textsc{Summary:} & \\
  14 &  \qquad \textsc{Synopsis of Muhammed's Treatise}   &   \pageref{par14} \\
  15 &  \qquad \textsc{Commandinus's Treatise}    &  \pageref{par15} \\
  16 &  \qquad \textsc{Synopsis of Euclid's Treatise}  &    \pageref{par16} \\
  17 &  \qquad \textsc{Analysis of Leonardo's Work}   &   \pageref{par17} \\
  &&\\  % blank line
\multicolumn{3}{c}{II} \\
18  &  \textsc{Abraham Savasorda, Jordanus Nemorarius}, & \\
    &  \textsc{Luca Paciuolo}   & \pageref{par18} \\
19  &  ``\textsc{Muhammed Bagdedinus}'' \textsc{and other} & \\
    & \textsc{Arabian writers on Divisions of Figures} & \pageref{par19} \\
20 &   \textsc{Practical Applications of the problems on Divisions} & \\
   & \textsc{of Figures; the}
   \ItalicGreekSnippet{metrik'a}
     \textsc{of Heron of Alexandria}   & \pageref{par20} \\
21  &  \textsc{Connection between Euclid's Book} \textit{On Divisions}, & \\
 &  \textsc{Apollonius's treatise} \textit{On Cutting off a Space} \textsc{and a} & \\
 & \textsc{Pappus-lemma to Euclid's book of} \textit{Porisms}    & \pageref{par21} \\
   &&\\  % blank line
\multicolumn{3}{c}{III} \\
22--57  &  \textsc{Restoration of Euclid's}
           \ItalicGreekSnippet{per`i diair'esewn bibl'ion} & \pageref{par22}\\
   &&\\  % blank line
\multicolumn{3}{c}{IV} \\
 & \textsc{Appendix}   & \pageref{appendix} \\
 & \textsc{Index of Names} & \pageref{indexpage}
\end{tabular}



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\mainmatter
\phantomsection\pdfbookmark[-1]{Main Matter}{Main Matter}
\pagestyle{fancy}
\Chapter{I}                                                               % Chapter I
\Section{\textit{Proclus, and Euclid's book On Divisions.}}
\Paragraph{PROCLUS, AND EUCLID'S BOOK ON DIVISIONS.}{1}                   % Paragraph 1
\phantomsection\pdfbookmark[0]{Proclus, and Euclid's book On Divisions.}{hyperPar1}
%%%%%%                              {Proclus, and Euclid's book On Divisions.}
Last in a list of Euclid's works ``full of admirable
diligence and skilful consideration,'' Proclus mentions, without
comment,
%[DPStyle Greek: peri diaireseôn biblion]
\ItalicGreekSnippet{per`i diair'esewn bibl'ion}\Footnote{1}{              % Footnote 1
\textit{Procli Diadochi in primum Euclidis elementorum
  librum commentarii} ex rec.
  G.~Friedlein, Leipzig, 1873, p.~69. Reference to this work will be
  made by ``Proclus.''}.                                                  % end footnote 1
But a little later\Footnote{2}{                                           % Footnote 2
Proclus\footnotemark[1], p.~144.}                                         % end footnote 2
in speaking of the conception or definition of \textit{figure} and of the
divisibility of a figure into others differing from it in kind,
Proclus adds: ``For the circle is divisible into parts unlike
in definition or notion, and so is each of the rectilineal figures;
this is in fact the business of the writer of the Elements in
his Divisions, where he divides given figures, in one case into
like figures, and in another into unlike\Footnote{3}{                     % Footnote 3
  In this translation I have followed \textsc{T.~L.\ Heath},
  \textit{The Thirteen Books of Euclid's Elements}, 1,
  Cambridge, 1908, p.~8. To Heath's account (pp.~8--10) of
  Euclid's book \textit{On Divisions} I shall refer by ``Heath.''

  ``Like'' and ``unlike'' in the above quotation mean,
  not ``similar'' and
  ``dissimilar'' in the technical sense,
  but ``like'' or ``unlike \emph{in definition} or \emph{notion}'':
  thus to divide a triangle into triangles would be to divide it
  into ``like'' figures, to
  divide a triangle into a triangle and a quadrilateral would be
  to divide it into ``unlike'' figures. (Heath.)}.''                      % end footnote 3



\Section{\textit{De Divisionibus by Muhammed Bagdedinus and the Dee MS.}}

\Paragraph{MSS. OF MUHAMMED BAGDEDINUS AND DEE}{2}                        % Paragraph 2
\phantomsection\pdfbookmark[0]{MSS of Muhammed Bagdedinus and Dee}{hyperPar2}
This is all we have from Greek sources, but the
discovery of an Arabian translation of the treatise supplies
the deficiency. In histories of Euclid's works (for example
% -----File: 014.png---- Folio 2 -----------------------------------------------------
those by Hankel\Footnote{4}{                                              % Footnote 4
  \textsc{H.~Hankel}, \textit{Zur Geschichte der Mathematik},
  Leipzig, 1874, p.~234.},                                                % end footnote 4
Heiberg\Footnote{5}{                                                      % Footnote 5
  \textsc{J.~L.~Heiberg},
  \textit{Litterargeschichtliche Studien über Euklid}, Leipzig, 1882,
  pp.~13--16, 36--38.
  Reference to this work will be made by ``Heiberg.''},                   % end footnote 5
Favaro\Footnote{6}{                                                       % Footnote 6
  \textsc{E.~A.~Favaro}.
  ``Preliminari ad una Restituzione del libro di Euclide sulla
  divisione delle figure piane,''
  \textit{Atti del reale Istituto Veneto di Scienze, Lettere ed Arti},
  \textsc{i}$_{6}$, 1883, pp.~393--6.
  ``Notizie storico-critiche sulla Divisione delee Aree''
  (Presentata li 28~gennaio, 1883),
  \textit{Memorie del reale Istituto Veneto di Scienze, Lettere ed Arti},
  \textsc{xxii}, 129--154.
  This is by far the most elaborate consideration of the subject up to
  the present. Reference to it will be made by ``Favaro.''
  },                                                                      % end footnote 6
Loria\Footnote{7}{                                                        % Footnote 7
  \textsc{G.~Loria},
  ``Le Scienze esatte nell' antica Grecia, Libro \textsc{ii},
  Il periodo aureo della geometria Greca.''
  \textit{Memorie della regia Accademia di Scienze,
  \DPtypo{Lettre}{Lettere} ed Arti in Modena},  \DPnote{Typo Corrected}
  \textsc{xi}$_{2}$, 1895, pp.~68--70, 220--221.
  \textit{Le Scienze esatte nell' antica Grecia},
  Seconda edizione. Milano, 1914, pp.~250--252, 426--427.},               % end footnote 7
Cantor\Footnote{8}{                                                       % Footnote 8
  \textsc{M.~Cantor},
  \textit{Vorlesungen über Geschichte der Mathematik},
  \textsc{i}$_{3}$, 1907, pp.~287--8;
  \textsc{ii}$_{2}$, 1900, p.~555.},                                      % end footnote 8
Hultsch\Footnote{9}{                                                      % Footnote 9
  \textsc{F.~Hultsch}, Article ``Eukleides'' in Pauly-Wissowa's
  \textit{Real-Encyclopädie der Class. Al\-tert\-ums\-wis\-sen\-schaf\-ten},
  \textsc{vi}, Stuttgart, 1909, especially Cols.~1040--41.},              % end footnote 9
Heath\footnotemark[3])
prominence is given to a treatise \textit{De Divisionibus}, by
one ``Muhammed Bagdedinus.'' Of this in 1563\Footnote{10}{                % Footnote 10
  When Dee was in Italy visiting Commandinus at Urbino.}                  % end footnote 10
a copy (in Latin) was given by John Dee to Commandinus who published
it in Dee's name and his own in 1570\Footnote{11}{                        % Footnote 11
  \textit{De superficierum divisionibus liber Machometo Bagdedino
  ascriptus nunc primum Joannis Dee Londinensis \& Federici Commandini
  Urbinatis opera in lucem editus}. Federici Commandini de eadem re
  libellus. Pisauri, \textsc{mdlxx}. In the same year appeared an
  Italian translation: \textit{Libro del modo di dividere le superficie
  attribuito a Machometo Bagdedino. Mandato in luce la prima volta da
  M.~G.\ Dee\ldots{}e da M.~F.\ldots Commandino\ldots{}Tradotti dal Latino
  in volgare da F.~Viani de' Malatesti,\ldots}
  In Pesaro, del \textsc{mdlxx}\ldots 4 unnumbered leaves
  and 44 numbered on one side.

  An English translation from the Latin, with the following title-page,
  was published in the next century:
  \textit{A Book of the Divisions of Superficies: ascribed
  to Machomet Bagdedine.
  Now put forth, by the pains of John Dee of London, and
  Frederic Commandine of Urbin.
  As also a little Book of Frederic Commandine,
  concerning the same matter.
  London Printed by R.~\& W.\ Leybourn}, 1660.
  Although this work has a separate title page and the above date,
  it occupies the last fifty pages (601--650) of a work dated a year
  later: \textit{Euclid's Elements of Geometry in XV Books\ldots{}to
  which is added a Treatise of Regular Solids by Campane and Flussas
  likewise Euclid's Data and Marinus Preface thereunto annexed.
  Also a Treatise of the Divisions of Superficies ascribed to Machomet
  Bagdedine, but published by Commandine, at the request of John Dee
  of London; whose Preface to the said Treatise declares it to be the
  Worke of Euclide, the Author of the Elements.
  Published by the care and Industry of John Leeke and George Serle,
  Students in the Mathematics}. London\ldots\textsc{mdclxi}.

  A reprint of simply that portion of the Latin edition which is the
  text of Muhammed's work appeared in:
  \ItalicGreekSnippet{EUKLEIDOU TA SWZOMENA}
  \textit{Euclidis quae supersunt omnia.
  Ex rescensione Davidis Gregorii}\ldots{}Oxoniae\ldots\textsc{mdcciii}.
  Pp.~665--684:
  \ItalicGreekSnippet{EUKLEIDOU WS OIONTAI TINES, PERI DIAIRESEWN BIBLOS}
  Euclidis, ut quidam arbitrantur,
  de divisionibus liber---vel ut alii volunt, Machometi Bagdedini
  liber de divisionibus superficierum.''}.                                % end footnote 11
Recent writers whose
publications appeared before 1905 have generally supposed
that Dee had somewhere discovered an Arabian original of
Muhammed's work and had given a Latin translation to
Commandinus. Nothing contrary to this is indeed explicitly
% -----File: 015.png---- Folio 3 -----------------------------------------------------
stated by Steinschneider when he writes in 1905\Footnote{12}{             % Footnote 12
  \textsc{M.~Steinschneider},
  ``Die Europäischen Übersetzungen aus dem Arabischen
  bis Mitte des 17.~Jahrhunderts.''
  \textit{Sitzungsberichte der Akademie der Wis\-sen\-schaf\-ten in Wien}
  (Philog.-histor.\ Klasse) \textsc{cli}, Jan.~1905, Wien, 1906.
  Concerning ``171.\ Muhammed'' \textit{cf}.~pp.~41--2.
  Reference to this paper will be made by ``Steinschneider.''},           % end footnote 12
``Machomet Bagdadinus (=aus Bagdad) heisst in einem alten MS.\
Cotton (jetzt im Brit.\ Mus.) der Verfasser von: de Superficierum
divisione (22~Lehrsätze); Jo.~Dee aus London entdeckte es
und übergab es T.~Commandino\ldots.'' For this suggestion as
to the place where Dee found the MS.\ Steinschneider gives
no authority. He does, however, give a reference to
Wenrich\Footnote{13}{                                                     % Footnote 13
  \textsc{J.~G.~Wenrich}, \textit{De auctorum Graecorum versionibus}.
  Lipsiae, \textsc{mdcccxlii}, p.~184.},                                  % end footnote 13
who in turn refers to a list of the printed books
(``Impressi'') of John Dee, in a life of Dee by Thomas
Smith\Footnote{14}{                                                       % Footnote 14
  \textsc{T.~Smith},
  \textit{Vitae quorundam eruditissimorum et illustrium virorum\ldots}
  {}Londini\ldots{}\textsc{mdccvii},
  p.~56. It was only the first 55 pages of this
  ``Vita Joannis Dee, Mathematici Angli,''
  which were translated into English by W.~A.\ Ayton, London, 1908.}      % end footnote 14
(1638--1710). We here find as the third in the list,
``Epistola ad eximium Ducis Urbini Mathematicum,
Fredericum Commandinum, praefixa libello Machometi Bagdedini
de superficierum divisionibus\ldots\textit{Pisauri}, 1570. Exstat
MS.\ in Bibliotheca Cottoniana sub Tiberio B \textsc{ix}.''

Then come the following somewhat mysterious sentences
which I give in translation\Footnote{15}{                                 % Footnote 15
  ``Post praefationem haec habet D.\ \textit{Usserius}
  Archiepiscopus Armachanus.
  \textit{Notandum est autem, Auctorem hunc} Euclide
  \textit{usum in Arabicam linguam converso, quem postea} Campanus
  \textit{Latinum fecit. Auctor igitur propositionum videtur fuisse}
  Euclides: \textit{demonstrationum, in quibus} Euclides
  \textit{in Arabico codice citatur},
  Machometus Bagded \textit{sive} Babylonius.''

  It has been stated that Campanus (13.~cent.) did not translate
  Euclid's Elements into Latin, but that the work published as his
  (Venice, 1482---the first printed edition of the \textit{Elements})
  was the translation made about 1120 by the English monk
  Athelhard of Bath. \textit{Cf.} \textsc{Heath},
  \textit{Thirteen Books of Euclid's Elements},
  \textsc{i}, 78, 93--96.}:                                               % end footnote 15
``After the preface Lord Ussher
[1581--1656], Archbishop of Armagh, has these lines: It is to
be noted that the author uses Euclid's Elements translated
into the Arabic tongue, which Campanus afterwards turned
into Latin. Euclid therefore seems to have been the author
of the Propositions [of \textit{De Divisionibus}] though not of the
demonstrations, which contain references to an Arabic edition
of the Elements, and which are due to Machometus of Bagded
or Babylon.'' This quotation from Smith is reproduced, with
various changes in punctuation and typography, by Kästner\Footnote{16}{   % Footnote 16
  \textsc{A.~G.~Kästner},
  \textit{Geschichte der Mathematik}\ldots{}
  Erster Band\ldots{}Göttingen, 1796,
  pp.~272--3. See also ``Zweyter'' Band, 1797, pp.~46--47.}.              % end footnote 16
Consideration of the latter part of it I shall postpone to a later
article (5).

% -----File: 016.png-----Folio 4----------------------------------------------------

\Paragraph{MSS. OF MUHAMMED BAGDEDINUS AND DEE}{3}                        % Paragraph 3
Following up the suggestion of Steinschneider, Suter
pointed out\Footnote{17}{                                                 % Footnote 17
  \textsc{H.~Suter},
  ``Zu dem Buche `De Superficierum divisionibus' des Muhammed Bagdedinus.''
  \textit{Bibliotheca Mathematica},
  \textsc{vi}$_3$, 321--2, 1905.},                                        % end footnote 17
without reference to Smith\footnotemark[14] or Kästner\footnotemark[16],
that in Smith's catalogue of the Cottonian Library there was an
entry\Footnote{18}{                                                       % Footnote 18
  \textsc{T.~Smith},
  \textit{Catalogus Librorum Manuscriptorum Bibliothecae Cottonianae\ldots}
  Oxonii,\ldots \textsc{mdcxcvi}, p.~24.}                                 % end footnote 18
under ``Tiberius\Footnote{19}{                                            % Footnote 19
  The original Cottonian library was contained in 14 presses,
  above each of which was a bust; 12 of these busts were of Roman
  Emperors. Hence the
  classification of the MSS.\ in the catalogue.}                          % end footnote 19
B \textsc{ix}, 6'': ``Liber Divisionum
Mahumeti Bag-dadini.'' As this MS.\ was undoubtedly in
Latin and as Cottonian MSS.\ are now in the British Museum,
Suter inferred that Dee simply made a copy of the above
mentioned MS.\ and that this MS.\ was now in the British
Museum. With his wonted carefulness of statement, Heath
does not commit himself to these views although he admits
their probable accuracy.

\Paragraph{MSS. OF MUHAMMED BAGDEDINUS AND DEE}{4}                        % Paragraph 4
As a final settlement of the question, I propose to
show that Steinschneider and Suter, and hence also many
earlier writers, have not considered all facts available. Some
of their conclusions are therefore untenable. In particular:

(1) In or before 1563 Dee did \emph{not} make a copy of any
Cottonian MS.;

(2) The above mentioned MS.\ (Tiberius, B.\ \textsc{ix}, 6) was
never, in its entirety, in the British Museum;

(3) The inference by Suter that this MS.\ was probably
the Latin translation of the tract from the Arabic, made by
Gherard of Cremona (1114--1187)---among the lists of whose
numerous translations a ``liber divisionum'' occurs---\emph{should
be accepted with great reserve};

(4) The MS.\ which Dee used can be stated with absolute
certainty and this MS.\ did not, in all probability, afterwards
become a Cottonian MS.

(1) Sir Robert Bruce Cotton, the founder of the Cottonian
Library, was born in 1571. The Cottonian Library was not,
therefore, in existence in 1563 and Dee could not then have
copied a Cottonian MS.

(2) The Cottonian Library passed into the care of the
nation shortly after 1700. In 1731 about 200 of the MSS.\
% -----File: 017.png---Folio 5------------------------------------------------------
were damaged or destroyed by fire. As a result of the
parliamentary inquiry Casley reported\Footnote{20}{                       % Footnote 20
  \textsc{D.~Casley}, p.~15\,ff.\ of
  \textit{A Report from the Committee appointed to view the
  Cottonian Library\ldots{}Published by order of the House of Commons}.
  London,
  \textsc{mdccxxxii} (British Museum MSS.\ 24932).
  \textit{Cf.}\ also the page opposite that numbered 120 in
  \textit{A Catalogue of the Manuscripts in the Cottonian Library\ldots{}with
  an Appendix containing an account of the damage sustained by the Fire in 1731;
  by S.~Hooper\ldots} London:\ldots\textsc{mdcclxxvii}.}                  % end footnote 20
on the MSS.\ destroyed
or injured. Concerning Tiberius \textsc{ix}, he wrote, ``This volume
burnt to a crust.'' He gives the title of each tract and the
folios occupied by each in the volume. ``Liber Divisionum
Mahumeti Bag-dadini'' occupied folios 254--258. When the
British Museum was opened in \emph{1753}, what was left of the
Cottonian Library was immediately placed there. Although
portions of all of the leaves of our tract are now to be seen
in the British Museum, practically none of the writing is
decipherable.

(3) Planta's catalogue\Footnote{21}{                                      % Footnote 21
  \textsc{J.~Planta},
  \textit{A Catalogue of the Manuscripts in the Cottonian Library
  deposited in the British Museum. Printed by command of his Majesty
  King George III\ldots} 1802.

  In the British Museum there are three MS.\ catalogues of the
  Cottonian Library:

  (1) \textit{Harleian MS.} 6018, a catalogue made in 1621.
  At the end are memoranda of loaned books.
  On a sheet of paper bearing date Novem.~23, 1638, Tiberius B
  \textsc{ix} is listed (folio~187) with its art.~4:
  ``liber divisione Machumeti Bagdedini.''
  The paper is torn so that the name of the person to whom the work
  was loaned is missing.
  The volume is not mentioned in the main catalogue.

  (2) \textit{MS.~No.}~36789, made after Sir Robert Cotton's death
  in 1631 and before 1638
  (\textit{cf.\ Catalogue of Additions to the MSS.\ in British Museum,
  1900--1905\ldots} London, 1907, pp.~226--227), contains,
  apparently, no reference to ``Muhammed.''

  (3) \textit{MS.~No.}~36682~A, of uncertain date but earlier than
  1654 (\textit{Catalogue of Additions\ldots{}l.c.}\ pp.~188--189).
  On folio 78 \textit{verso} we find Tiberius B \textsc{ix}, Art.\ 4:
  ``Liber divisione Machumeti Bagdedini.''

  A ``Muhammed'' MS.\ was therefore in the Cottonian Library in 1638.

  The anonymously printed (1840?) ``Index to articles printed from
  the Cotton MSS., \& where they may be found'' which may be seen
  in the British Museum, only gives references to the MSS.\
  in ``Julius.''}                                                         % end footnote 21
has the following note concerning
Tiberius \textsc{ix}: ``A volume on parchment, which once consisted
of 272 leaves, written about the XIV.\ century [not the
XII.\ century, when Gherard of Cremona flourished], containing
eight tracts, the principal of which was a `Register
of William Cratfield, abbot of St Edmund'\,'' [d.~1415].
Tracts 3, 4, 5 were on music.

(4) On ``$A^\circ$ 1583, 6 Sept.''\ Dee made a catalogue of the
MSS.\ which he owned.  This catalogue, which is in the
Library of Trinity College, Cambridge\Footnote{22}{                       % Footnote 22
  A transcription of the Trinity College copy, by Ashmole,
  is in MS.\ Ashm.\ 1142. Another autograph copy is in the
  British Museum: Harleian MS.\ 1879.},                                   % end footnote 22
has been published\Footnote{23}{                                          % Footnote 23
  \textit{Camden Society Publications},
  \textsc{xix}, London, \textsc{m.dccc.xlii}.}                            % end footnote 23
% -----File: 018.png---Folio 6------------------------------------------------------
under the editorship of J.~O.~Halliwell. The 95th item
described is a folio parchment volume containing 24 tracts
on mathematics and astronomy. The 17th tract is entitled
``Machumeti Bagdedini liber divisionum.'' As the contents
of this volume are entirely different from those of Tiberius \textsc{ix}
described above, in~(3), it seems probable that there were two
copies of ``Muhammed's'' tract, while the MS.\ which Dee
used for the 1570 publication was undoubtedly his own, as we
shall presently see. If the two copies be granted, there is no
evidence against the Dee copy having been that made by
Gherard of Cremona.

\Paragraph{MSS. OF MUHAMMED BAGDEDINUS AND DEE}{5}                        % Paragraph 5
There is the not remote possibility that the Dee MS.\
was destroyed soon after it was catalogued. For in the same
month that the above catalogue was prepared, Dee left his
home at Mortlake, Surrey, for a lengthy trip in Europe.
Immediately after his departure ``the mob, who execrated
him as a magician, broke into his house and destroyed a great
part of his furniture and books\Footnote{24}{                             % Footnote 24
  \textit{Dictionary of National Biography}, Article,
  ``Dee, John.''}\ldots''\                                                % end footnote 24
many of which ``were the
written bookes\Footnote{25}{                                              % Footnote 25
  ``The compendious rehearsall of John Dee his dutifull declaration A.~1592''
  printed in \textit{Chetham Miscellanies},
  vol.~\textsc{i}, Manchester, \textsc{1851}, p.~27.}.''                  % end footnote 25
Now the Dee catalogue of his MSS.\
(MS.\ O.\ iv.~20), in Trinity College Library, has numerous
annotations\Footnote{26}{                                                 % Footnote 26
  Although Halliwell professed to publish the Trinity MS., he makes not
  the slightest reference to these annotations.}                          % end footnote 26
in Dee's handwriting. They indicate just what
works were (1) destroyed or stolen (``Fr.'')\Footnote{27}{                % Footnote 27
  ``Fr.''\ is no doubt an abbreviation for \textit{Furatum}.}\            % end footnote 27
and (2) left(``T.'')\Footnote{28}{                                        % Footnote 28
  ``T.'', according to Ainsworth (\textit{Latin Dictionary}),
  was put after the name of a
  soldier to indicate that he had survived (\textit{superstes}).
  Whence this abbreviation?}                                              % end footnote 28
after the raid. Opposite the titles of the tracts in the volume
including the tract ``liber divisionum,'' ``Fr.''\ is written, and
opposite the title ``Machumeti Bagdedini liber divisionum''
is the following note: ``Curavi imprimi Urbini in Italia per
Federicum Commandinum exemplari descripto ex vetusto isto
monumento(?) per me ipsum.'' Hence, as stated above, it is
now definitely known (1) that the MS.\ which Dee used was
his own, and (2) that some 20 years after he made a copy, the
MS.\ was stolen and probably destroyed\Footnote{29}{                       % Footnote 29
  The view concerning the theft or destruction of the MS.\ is borne
  out by the fact that in a catalogue of Dee's Library
  (British Museum MS.~35213) made early in the seventeenth century
  (\textit{Catalogue of Additions and Manuscripts}\ldots{}1901,
  p.~211), Machumeti Bagdedini is not mentioned.}.                         % end footnote 29

On the other hand we have the apparently contradictory
% -----File: 019.png---Folio 7------------------------------------------------------
evidence in the passage quoted above (Art.~2) from the life
of Dee by Smith\footnotemark[14]
who was also the compiler of the Catalogue
of the Cottonian Library. Smith was librarian when he
wrote both of these works, so that any definite statement
which he makes concerning the library long in his charge is
not likely to be successfully challenged. Smith does not
however say that Dee's ``Muhammed'' MS.\ was in the
Cottonian Library, and if he knew that such was the case
we should certainly expect some note to that effect in the
catalogue\footnotemark[18];
for in three other places in his catalogue
(Vespasian B \textsc{x}, A \textsc{ii}$_{13}$,
Galba E \textsc{viii}), Dee's original
ownership of MSS.\ which finally came to the Cottonian
Library is carefully remarked. Smith does declare, however,
that the Cottonian MS.\ bore, ``after the preface,'' certain
notes (which I have quoted above) by Archbishop Ussher
(1581--1656). Now it is not a little curious that these notes
by Ussher, who was not born till after the Dee book was
printed, should be practically identical with notes in the
printed work, just after Dee's letter to Commandinus (Art.~3).
For the sake of comparison I quote the notes in question\Footnote{30}{    % Footnote 30
  This quotation from the Leeke-Serle Euclid\footnotemark[11]
  is an exact translation of the original.};                              % end footnote 30
\DPnote{The following quote contains instances of the long-s.}
\label{longs}
``To the Reader.---I am here to advertise thee (kinde Reader)
that this author which we present to thee, made use of Euclid
translated into the Arabick Tongue, whom afterwards Campanus
made to speake Latine. This I thought fit to tell thee, that
so in searching or examining the Propositions which are cited
by him, thou mightest not sometime or other trouble thy selfe
in vain, Farewell.''

The Dee MS.\ as published did not have any preface.
We can therefore only assume that Ussher wrote in a MS.\
which \emph{did} have a preface the few lines which he may have
seen in Dee's printed book.

\Paragraph{MSS. OF MUHAMMED BAGDEDINUS AND DEE}{6}                        % Paragraph 6
Other suggestions which have been made concerning
``Muhammed's'' tract should be considered. Steinschneider
asks, ``Ob identisch de Curvis superficiebus, von einem
Muhammed, MS.\ Brit.\ Mus.\ Harl.\ 623\footnotemark[6]
(\textsc{i}, 191)\Footnote{31}{                                           % Footnote 31
  This should be 625\footnotemark[6]
  (\textsc{i}, 391).}?''                                                  % end footnote 31
I have examined this MS.\ and found that it has nothing to do with
the subject matter of the Dee tract.

But again, Favaro states\Footnote{32}{                                    % Footnote 32
  Favaro, p.~140. \textit{Cf.}\ Heiberg, p.~14.
  This suggestion doubtless originated
  with Ofterdinger\footnotemark[38],
p.~[1].}:                                                                 % end footnote 32
``Probabilmente il manoscritto
% -----File: 020.png---Folio 8------------------------------------------------------
del quale si \DPtypo{servi}{servì}\DPnote{[**sic but should be servì]. Typo Corrected}
il Dee è lo stesso indicato dall'Heilbronner\Footnote{33}{                % Footnote 33
  \textsc{J.~C.~Heilbronner},
  \textit{Historia matheseos Universae\ldots}Lipsiae, \textsc{mdccxlii},
  p.~620: (``Manuscripta mathematica in Bibliotheca Bodlejana'')
  ``34 Mohammedis
  Bagdadeni liber de superficierum divisionibus, cum Notis H.~S.''}       % end footnote 33
\DPtypo{comme}{come}\DPnote{[**sic but should be come] Typo Corrected}
esistente nella Biblioteca Bodleiana di Oxford.'' Under
date ``6.~3.~1912'' Dr A.~Cowley, assistant librarian in the
Bodleian, wrote me as follows: ``We do not possess a copy
of Heilbronner's Hist.\ Math.\ Univ. In the old catalogue of
MSS.\ which he would have used, the work you mention is included---but
is really a printed book and is only included in the
catalogue of MSS.\ because it contains some manuscript notes---

``Its shelf-mark is Savile T 20.

``It has 76 pages in excellent condition. The title page
has: De Superficierum $|$ divisionibus liber $|$ Machometo Bagdedino $|$
ascriptus $|$ nunc primum Joannis Dee $|$ \ldots $|$ opera in
lucem editus $|$ \ldots{}Pisauri \textsc{mdlxx}.

``The MS.\ notes are by Savile, from whom we got the
collection to which this volume belongs.''

The notes were incorporated into the Gregory edition\footnotemark[11]
of the Dee tract. Here and elsewhere\Footnote{34}{                        % Footnote 34
  \textsc{H.~Savile}, \textit{Praelectiones tresdecim in principium
  elementorum Evclidis, Oxonii habitae M.DC.XX}. Oxonii\ldots,
  1621, pp.~17--18.}                                                      % end footnote 34
Savile objected to
attributing the tract to Euclid as author\Footnote{35}{                   % Footnote 35
  Dee's statement of the case in his letter to Commandinus
  (Leeke-Serle Euclid\footnotemark[11], \textit{cf.}\ note~30)
  is as follows: ``As for the authors name, I would have you
  understand, that to the very old Copy from whence I writ it,
  the name of \textsc{Machomet Bagdedine} was put in ziphers or
  Characters, (as they call them) who whether he were that
  \textit{Albategnus} whom \textit{Copernicus} often cites as a very
  considerable Author in Astronomie; or that Machomet who is said to
  have been \textit{Alkindus's} scholar, and is reported to have
  written somewhat of the art of Demonstration, I am not yet certain of:
  or rather that this may be deemed a Book of our \textit{Euclide},
  all whose Books were long since turned out of the Greeke into the
  Syriack and Arabick Tongues. Whereupon, It being found some time or
  other to want its Title with the \textit{Arabians} or
  \textit{Syrians}, was easily attributed by the transcribers
  to that most famous Mathematician among them, Machomet:
  which I am able to prove by many testimonies, to be often done in
  many Moniments of the Ancients;
  \ldots{}yea further, we could not yet perceive so great acuteness of
  any \textit{Machomet} in the Mathematicks, from their moniments
  which we enjoy, as everywhere appears in these Problems.
  Moreover, that \textit{Euclide} also himself wrote one Book
  \ItalicGreekSnippet{peri diair'esewn}
  that is to say, \textit{of Divisions}, as may be evidenced from
  Proclus's Commentaries upon his first of \textit{Elements}:
  and we know none other extant under this title, nor can we find any,
  which for excellencie of its treatment, may more rightfully or
  worthily be ascribed to \textit{Euclid}. Finally, I remember that in
  a certain very ancient piece of Geometry, I have read a place cited
  out of this little Book in expresse words, even as from amost
  (\textit{sic}) certain work of \textit{Euclid}. Therefore we have
  thus briefly declared our opinions for the present, which we desire
  may carry with them so much weight, as they have truth in
  them\ldots. But whatsoever that Book of \textit{Euclid} was
  concerning Divisions, certainly this is such an one as may be both
  very profitable for the studies of many, and also bring much honour
  and renown to every most noble ancient Mathematician; for the most
  excellent acutenesse of the invention, and the most accurate
  discussing of all the Cases in each Probleme\ldots.''}.               % end footnote 35
His arguments
% -----File: 021.png---Folio 9------------------------------------------------------
are summed up, for the most part, in the conclusions of
Heiberg followed by Heath: ``the Arabic original could not
have been a direct translation from Euclid, and probably was
not even a direct adaptation of it; it contains mistakes and
unmathematical expressions, and moreover does not contain
the propositions about the division of a circle alluded to by
Proclus. Hence it can scarcely have contained more than
a fragment of Euclid's work.''

\Section{\textit{The Woepcke-Euclid MS.}}
\Paragraph{THE WOEPCKE-EUCLID MS.}{7}                                     % Paragraph 7
\phantomsection\pdfbookmark[0]{The Woepcke-Euclid MS.}{hyperPar7}
On the other hand Woepcke found in a MS.\ (No.~952.~2
Arab.\ Suppl.)\ of the Bibliothèque nationale, Paris, a treatise
in Arabic on the division of plane figures, which he translated,
and published in 1851\Footnote{36}{                                       % Footnote 36
  \textsc{F.~Woepcke},
  ``Notice sur des traductions Arabes de deux ouverages perdus d'Euclide''
  \textit{Journal Asiatique}, Septembre--Octobre, 1851,
  \textsc{xviii}$_4$, 217--247.
  Euclid's work \textit{On the division (of plane figures)}: pp.~233--244.
  Reference to this paper will be made by ``Woepcke.''
  In \textit{Euclidis opera omnia}, vol.~8, now in the press, there are
  ``Fragmenta collegit et disposuit J.~L.\ Heiberg,'' through whose
  great courtesy I have been enabled to see the proof-sheets.
  First among the fragments, on pages
  227--235, are (1) the Proclus references to
  \ItalicGreekSnippet{peri diair'esewn}
and (2) the Woepcke
  translation mentioned above. In the article on Euclid in the last edition of the
  \textit{Encyclopaedia Britannica} no reference is made to this work or to the writings
  of Heiberg, Hultsch, Steinschneider and Suter.}.                        % end footnote 36
``It is expressly attributed to Euclid
in the MS.\ and corresponds to the description of it by Proclus.
Generally speaking, the divisions are divisions into figures of
the same kind as the original figures, e.~g.\ of triangles into
triangles; but there are also divisions into `unlike' figures,
e.~g.\ that of a triangle by a straight line parallel to the base.
The missing propositions about the division of a circle are
also here: `to divide into two equal parts a given figure
bounded by an arc of a circle and two straight lines including
a given angle' and `to draw in a given circle two parallel
straight lines cutting off a certain part of a circle.' Unfortunately
the proofs are given of only four propositions (including the
two last mentioned) out of 36, because the Arabian translator
found them too easy and omitted them.'' That the omission
is due to the translator and did not occur in the original is
indicated in two ways, as Heiberg points out. Five auxiliary
propositions (Woepcke 21, 22, 23, 24, 25) of which no use is
made are introduced. Also Woepcke 5 is: ``\ldots{}and we divide
the triangle by a construction analogous to the preceding
construction''; but no such construction is given.

The four proofs that are given are elegant and depend
% -----File: 022.png---Foilio 10------------------------------------------------------
only on the propositions (or easy deductions from them) of
the \textit{Elements}, while Woepcke~18 has the true Greek ring:
``to apply to a straight line a rectangle equal to the rectangle
contained by $\mathit{AB}$, $\mathit{AC}$ and \emph{deficient by a square}.''

\Paragraph{THE WOEPCKE-EUCLID MS.}{8}                                     % Paragraph 8

To no proposition in the Dee MS.\ is there word for
word correspondence with the propositions of Woepcke but
in content there are several cases of likeness. Thus, Heiberg
continues,
\begin{verse}
  Dee 3 = Woepcke 30 (a special case is Woepcke 1);\\
  Dee 7 = Woepcke 34 (a special case is Woepcke 14);\\
  Dee 9 = Woepcke 36 (a special case is Woepcke 16);\\
  Dee 12 = Woepcke 32 (a special case is Woepcke 4).
\end{verse}

Woepcke~3 is only a special case of Dee~2; Woepcke~6,
7, 8, 9 are easily solved by Dee~8. And it can hardly be
chance that the proofs of exactly these propositions in Dee
should be without fault. That the treatise published by
Woepcke is no fragment but the complete work which was
before the translator is expressly stated\Footnote{37}{                   % Footnote 37
  Woepcke, p.~244.},                                                      % end footnote 37
``fin du traité.'' It
is moreover a well ordered and compact whole. Hence we
may safely conclude that Woepcke's is not only Euclid's own
work but the whole of it, except for proofs of some propositions.

\Paragraph{THE WOEPCKE-EUCLID MS.}{9}                                     % Paragraph 9
For the reason just stated the so-called
\textit{Wiederherstellung}
of Euclid's work by Ofterdinger\Footnote{38}{                             % Footnote 38
  \textsc{L.~F.~Ofterdinger},
  \textit{Beiträge zur Wiederherstellung der Schrift des Euklides
  über der Theilung der Figuren}, Ulm, 1853.},                            % end footnote 38
based mainly on
Dee, is decidedly misnamed. A more accurate description of
this pamphlet would be, ``A translation of the Dee tract with
indications in notes of a certain correspondence with 15 of
Woepcke's propositions, the whole concluding with a translation
of the enunciations of 16 of the remaining 21 propositions
of Woepcke not previously mentioned.'' Woepcke~30, 31, 34,
35, 36 are not even noticed by Ofterdinger. Hence the claim
I made above (``Introductory'') that the first real restoration
of Euclid's work is now presented. Having introduced
Woepcke's text as one part of the basis of this restoration,
the other part demands the consideration of the

\Section{\textit{Practica Geometriae of Leonardo Pisano (Fibonaci).}}

\Paragraph{PRACTICA GEOMETRIAE OF LEONARDO PISANO}{10}                    % Paragraph 10
\phantomsection\pdfbookmark[0]{Practica Geometriae of Leonardo Pisano}{hyperPar10}
It was in the year 1220 that Leonardo Pisano, who
occupies such an important place in the history of mathematics
% -----File: 023.png---Folio 11------------------------------------------------------
of the thirteenth century\Footnote{39}{                                   % Footnote 39
  \textsc{M.~Cantor},
  \textit{Vorlesungen über Geschichte der Mathematik},
  \textsc{ii}$_{2}$, 1900, pp.~3--53;
  ``Practica Geometriae,'' pp.~35--40.},                                  % end footnote 39
wrote his \textit{Practica Geometriae}, and
the MS.\ is now in the Vatican Library. Although it was
known and used by other writers, nearly six and one half
centuries elapsed before it was finally published by Prince
Boncompagni\Footnote{40}{                                                 % Footnote 40
  \textit{Scritti di Leonardo Pisano matematico del secolo decimoterzo
  \DPtypo{publicati}{pubblicati}
  da \DPtypo{Baldasarre}{Baldassarre}
%% [**sic but should be Baldassarre -- see books.google.com]
  Boncompagni}.  Volume \textsc{ii}
  (Leonardi Pisani Practica Geometriae ed opuscoli). Roma\ldots{}1862.
  Practica Geometriae, pp.~1--224.}.                                      % end footnote 40
Favaro was the first\footnotemark[6]
to call attention to
the importance of Section IIII\Footnote{41}{                              % Footnote 41
  \textit{Scritti di Leonardo Pisano\ldots}\textsc{ii}, pp.~110--148.}    % end footnote 41
of the \textit{Practica Geometriae} in
connection with the history of Euclid's work. This section
is wholly devoted to the enunciation and proof and numerical
exemplification of propositions concerning the divisions of
figures. Favaro reproduces the enunciations of the propositions
and numbers them 1 to 57\Footnote{42}{                                    % Footnote 42
  These numbers I shall use in what follows. Favaro omits some auxiliary
  propositions and makes slips in connection with 28 and 40.
  Either 28 should have been more general in statement or another number
  should have been introduced.
  Similarly for 40. Compare Articles 33--34, 35.}.                        % end footnote 42
He points out that in both
enunciation and proof Leonardo 3, 10, 51, 57 are identical
with Woepcke 19, 20, 29, 28 respectively. But considerably
more remains to be remarked.

\Paragraph{PRACTICA GEOMETRIAE OF LEONARDO PISANO}{11}                    % Paragraph 11
No less than twenty-two of Woepcke's propositions
are practically identical in statement with propositions in
Leonardo; the solutions of eight more of Woepcke are either
given or clearly indicated by Leonardo's methods, and all six
of the remaining Woepcke propositions (which are auxiliary)
are assumed as known in the proofs which Leonardo gives of
propositions in Woepcke. Indeed, these two works have
a remarkable similarity. Not only are practically all of the
Woepcke propositions in Leonardo, but the proofs called
for by the order of the propositions and by the auxiliary
propositions in Woepcke are, with a possible single exception\footnotemark[91],
invariably the kind of proofs which Euclid might have given---no
other propositions but those which had gone before or
which were to be found in the \textit{Elements} being required in
the successive constructions.

Leonardo had a wide range of knowledge concerning
Arabian mathematics and the mathematics of antiquity. His
\textit{Practica Geometriae} contains many references to Euclid's
\textit{Elements}
and many uncredited extracts from this work\Footnote{43}{                 % Footnote 43
  For example, on pages 15--16, 38, 95, 100--1, 154.}.                    % end footnote 43
% -----File: 024.png---Folio 12------------------------------------------------------
Similar treatment is accorded works of other writers. But
in the great elegance, finish and rigour of the whole,
originality of treatment is not infrequently evident. If
Gherard of Cremona made a translation of Euclid's book
\textit{On Divisions}, it is not at all impossible that this may have
been used by Leonardo. At any rate the conclusion seems
inevitable that he must have had access to some such MS.\ of
Greek or Arabian origin.

Further evidence that Leonardo's work was of Greek-Arabic
extraction can be found in the fact that, in connection
with the 113 figures, of the section On Divisions, of Leonardo's
work, the lettering in only 58 contains the letters \textit{c} or \textit{f}; that
is, the Greek-Arabic succession \textit{a, b, g, d, e, z \ldots} is used almost
as frequently as the Latin \textit{a, b, c, d, e, f, g,\ldots}; elimination of
Latin letters added to a Greek succession in a figure, for the
purpose of numerical examples (in which the work abounds),
makes the balance equal.

\Paragraph{PRACTICA GEOMETRIAE OF LEONARDO PISANO}{12}                    % Paragraph 12
My method of restoration of Euclid's work has been
as follows. Everything in Woepcke's text (together with
his notes) has been translated literally, reproduced without
change and enclosed by quotation marks. To all of Euclid's
enunciations (unaccompanied by constructions) which corresponded
to enunciations by Leonardo, I have reproduced
Leonardo's constructions and proofs, with the same lettering
of the figures\Footnote{44}{                                              % Footnote 44
  This is done in order to give indication of the possible origin of the construction
  in question (Art.~11).},                                                % end footnote 44
but occasional abbreviation in the form of
statement; that is, the extended form of Euclid in Woepcke's
text, which is also employed by Leonardo, has been sometimes
abridged by modern notation or briefer statement. Occasionally
some very obvious steps taken by Leonardo have been left
out but all such places are clearly indicated by explanation
in square brackets, [\quad]. Unless stated to the contrary, and
indicated by different type, no step is given in a construction
or proof which is not contained in Leonardo. When there
is no correspondence between Woepcke and Leonardo I have
exercised care to reproduce Leonardo's methods in other propositions,
as closely as possible. If, in a given proposition,
the method is extremely obvious on account of what has gone
before, I have sometimes given little more than an indication
of the propositions containing the essence of the required
% -----File: 025.png---Folio 13------------------------------------------------------
construction and proof. In the case of the six auxiliary
propositions, the proofs supplied seemed to be readily suggested
by propositions in Euclid's \textit{Elements}.

\Paragraph{PRACTICA GEOMETRIAE OF LEONARDO PISANO}{13}                    % Paragraph 13
Immediately after the enunciations of Euclid's
problems follow the statements of the correspondence with
Leonardo; if exact, a bracket encloses the number of the
Leonardo proposition, according to Favaro's numbering, and
the page and lines of Boncompagni's edition where Leonardo
enunciates the same proposition.

The following is a comparative table of the Euclid and, in
brackets, of the corresponding Leonardo problems: 1~(5);
2~(14); 3~(2,~1); 4~(23); 5~(33); 6~(16); 7~(20)\Footnote{45}{            % Footnote 45
  Leonardo considers the case of ``one third'' instead of Euclid's
  ``a certain fraction,'' but in the case of 20 he concludes that in
  the same way the figure may be divided
  ``into four or many equal parts.''
  \textit{Cf.}\ Article~28.};                                             % end footnote 45
8 (27)\Footnote{46}{                                                      % Footnote 46
  Woepcke~8 may be considered as a part of Leonardo~27 or better as an
  unnumbered proposition following Leonardo~25.};                         % end footnote 46
9 (30, 31)\Footnote{47}{                                                  % Footnote 47
  Leonardo's propositions 30--32 consider somewhat more general problems
  than Euclid's~9 and 13. \textit{Cf.}\ Articles~30 and 34.};             % end footnote 47
10~(18); 11~(0); 12~(28)\footnotemark[42];
13~(32)\footnotemark[47]; 14~(36);
15~(40); 16~(37); 17~(39); 18~(0); 19~(3); 20~(10); 21~(0);
22~(0); 23~(0); 24~(0); 25~(0); 26~(4); 27~(11); 28~(57);
29 (51)\footnotemark[45];
30~(0); 31~(0); 32~(29); 33~(35); 34~(40)\footnotemark[42];
35~(0);
36~(0).

\Section{\textit{Summary}}

It will be instructive, as a means of comparison, to set
forth in synoptic fashion: (1) the Muhammed-Commandinus
treatise; (2) the Euclid treatise; (3) Leonardo's work. In
(1) and (2) I follow Woepcke closely\Footnote{48}{                        % Footnote 48
  Woepcke, pp.~245--246.}.                                                % end footnote 48

\medskip

\Paragraph{SYNOPSIS OF MUHAMMED'S TREATISE}{14}                           % Paragraph 14
\phantomsection\pdfbookmark[0]{Summary}
                              {Summary}
\phantomsection\pdfbookmark[1]{Synopsis of Muhammed's Treatise}{hyperPar14}
\textit{Synopsis of Muhammed's Treatise---}
\renewcommand{\labelenumi}{\Roman{enumi}.}
\renewcommand{\labelenumii}{\Alph{enumii}.}
\renewcommand{\labelenumiii}{\arabic{enumiii}.}
\begin{enumerate}
\item In all the problems it is required to divide the proposed
figure into two parts having a given ratio.
\item The figures divided are: the triangle (props.~1--6);
the parallelogram (11); the trapezium\footnotemark[89]
(8, 12, 13); the quadrilateral
(7, 9, 14--16); the pentagon (17, 18, 22); a pentagon
with two parallel sides (19), a pentagon of which a side is
parallel to a diagonal (20).
% -----File: 026.png---Folio 14------------------------------------------------------
\item The transversal required to be drawn:
\begin{enumerate}
\item passes through a given point and is situated:
\begin{enumerate}
\item at a vertex of the proposed figure (1, 7, 17);
\item on any side (2, 9, 18);
\item on one of the two parallel sides (8).
\end{enumerate}

\item is parallel:
\begin{enumerate}
\item to a side (not parallel) (3, 13, 14, 22);
\item to the parallel sides (11, 12, 19);
\item to a diagonal (15, 20);
\item to a perpendicular drawn from a vertex of the
figure to the opposite side (4);
\item to a transversal which passes through a vertex of the figure (5);
\item to any transversal (6, 16).
\end{enumerate}
\end{enumerate}
\item Prop.~10: Being given the segment $\mathit{AB}$ and two
lines which pass through the extremities of this segment and
form with the line $\mathit{AB}$ any angles, draw a line parallel to $AB$
from one or the other side of $\mathit{AB}$ and such as to produce
a trapezium of given size.

Prop.~21. Auxiliary theorem regarding the pentagon.
\end{enumerate}

\bigskip
\Paragraph{COMMANDINUS'S TREATISE}{15}                                    % Paragraph 15
\phantomsection\pdfbookmark[1]{Commandinus's Treatise}{hyperPar15}
\textit{Commandinus's Treatise}---Appended to the first
published edition of Muhammed's work was a short treatise\Footnote{49}{   % Footnote 49
  Commandinus\footnotemark[11], pp.~54--76.}                              % end footnote 49
by Commandinus who said\Footnote{50}{                                     % Footnote 50
  Commandinus\footnotemark[11], p.~[ii]; Leeke-Serle Euclid, p.~603.}     % end footnote 50
of Muhammed: ``for what things
the author of the book hath at large comprehended in many
problems, I have compendiously comprised and dispatched in
two only.'' This statement repeated by Ofterdinger\Footnote{51}{          % Footnote 51
  Ofterdinger\footnotemark[38], p.~11, note.}                             % end footnote 51
and Favaro\Footnote{52}{                                                  % Footnote 52
  Favaro\footnotemark[6], p.~139.}                                        % end footnote 52
is somewhat misleading.

The ``two problems'' of Commandinus are as follows:

``Problem I\@. To divide a right lined figure according to
a proportion given, from a point given in any part of the
ambitus or circuit thereof, whether the said point be taken in
any angle or side of the figure.''

``Problem II\@. To divide a right lined figure $\mathit{GABC}$,
% -----File: 027.png---Folio 15------------------------------------------------------
according to a proportion given, $E$ to $F$, by a right line
parallel to another given line $D$.''

But the first problem is divided into 18 cases: 4 for the
triangle, 6 for the quadrilateral, 4 for the pentagon, 2 for the
hexagon and 2 for the heptagon; and the second problem, as
Commandinus treats it, has 20 cases: 3 for the triangle, 7 for
the quadrilateral, 4 for the pentagon, 4 for the hexagon, 2 for
the heptagon.

\Paragraph{SYNOPSIS OF EUCLID'S TREATISE}{16}                             % Paragraph 16
\phantomsection\pdfbookmark[1]{Synopsis of Euclid's Treatise}{hyperPar16}
%%%%%%                              {Synopsis of Euclid's Treatise}
\textit{Synopsis of Euclid's Treatise---}
\renewcommand{\labelenumi}{\Roman{enumi}.}
\renewcommand{\labelenumii}{\arabic{enumii}.}
\renewcommand{\labelenumiii}{\arabic{enumiii}.}

\begin{enumerate}
\item The proposed figure is divided:
\begin{enumerate}
\item into two equal parts (1, 3, 4, 6, 8, 10, 12, 14, 16,
19, 26, 28);
\item into several equal parts (2, 5, 7, 9, 11, 13, 15, 17, 29);
\item into two parts, in a given ratio (20, 27, 30, 32, 34, 36);
\item into several parts, in a given ratio (31, 33, 35, 36).
\end{enumerate}

The construction 1 or 3 is always followed by the construction
of 2 or 4, except in the propositions 3, 28, 29.

\item The figures divided are:
\renewcommand{\labelenumii}{ }
\begin{enumerate}
\item the triangle (1, 2, 3, 19, 20, 26, 27, 30, 31);
\item the parallelogram (6, 7, 10, 11);
\item the trapezium (4, 5, 8, 9, 12, 13, 32, 33);
\item the quadrilateral (14, 15, 16, 17, 34, 35, 36);
\item a figure bounded by an arc of a circle and two lines (28);
\item the circle (29).
\end{enumerate}

\renewcommand{\labelenumi}{\Roman{enumi}.}
\renewcommand{\labelenumii}{\Alph{enumii}.}
\renewcommand{\labelenumiii}{\arabic{enumiii}.}
%\renewcommand{\labelenumiiii}{\arabic{enumiiii}.}

\item It is required to draw a transversal:
\begin{enumerate}
\item passing through a point situated:
\begin{enumerate}
\item at a vertex of the figure (14, 15, 34, 35);
\item on any side (3, 6, 7, 16, 17, 36);
\item on one of two parallel sides (8, 9);
\item at the middle of the arc of the circle (28);
\item in the interior of the figure (19, 20);
\item outside the figure (10, 11, 26, 27);
\item in a certain part of the plane of the figure (12, 13)
\end{enumerate}

% -----File: 028.png---Folio 16------------------------------------------------------
\item parallel to the base of the proposed figure (1, 2,
4, 5, 30--33).

\item parallel to one another, the problem is indeterminate (29).
\end{enumerate}
\item Auxiliary propositions:

18.~To apply to a given line a rectangle of given size
and deficient by a square.

21, 22, when $a\centerdot d \gtrless b\centerdot c$, it follows that $a: b\gtrless c: d$;

%$x\gtrless y  z \lessgtr q$

23, 24, when $a:b > c:d$, it follows that
\[ (a \mp b) :b > (c \mp d): d;\]

25, when $a: b < c: d$, it follows that $(a-b): b<(c-d): d$.
\end{enumerate}

In the synopsis of the last five propositions I have
changed the original notation slightly.

\Paragraph{ANALYSIS OF LEONARDO'S WORK}{17}                               % Paragraph 17
\phantomsection\pdfbookmark[1]{Analysis of Leonardo's Work}{hyperPar17}
%%%%%%                              {Analysis of Leonardo's Work}
\textit{Analysis of Leonardo's Work}. I have not thought
it necessary to introduce into this analysis the unnumbered
propositions referred to above\footnotemark[42].

\begin{enumerate}
\item The proposed figure is divided:
\begin{enumerate}
\item into two equal parts (1--5, 15--18, 23--28, 36--38,
42--46, 53--55, 57);

\item into several equal parts (6, 7, 9, 13, 14, 19, 21,
33, 47--50, 56);

\item into two parts in a given ratio (8, 10--12, 20, 29--32,
 34, 39, 40, 51, 52);

\item into several parts in a given ratio (22, 35, 41).
\end{enumerate}

The construction 1 or 3 is always followed by the
construction of 2 or 4 except in the propositions 42--46,
51, 54, 57.

\item The figures divided are:

the triangle (1--14);

the parallelogram (15--22);

the trapezium (23--35);

the quadrilateral (36--41);

the pentagon (42--43);

the hexagon (44);

the circle and semicircle (45--56);

a figure bounded by an arc of a circle and two lines (57).

% -----File: 029.png---Folio 17------------------------------------------------------
\renewcommand{\labelenumii}{(\roman{enumii})}
\renewcommand{\labelenumiii}{\Alph{enumiii}.}
\renewcommand{\labelenumiv}{\arabic{enumiv}.}

\item \;
\begin{enumerate}
\item It is required to draw a transversal:
\begin{enumerate}
\item passing through a point situated:
\begin{enumerate}
\item at a vertex of the figure (1, 6, 26, 31, 34, 36, 41--44);
\item on a side not produced (2, 7, 8, 16, 20, 37, 39);
\item at a vertex or a point in a side (40);
\item on one of two parallel sides (24, 25, 27, 30);
\item on the middle of the arc of the circle (53, 55, 57);
\item on the circumference or outside of the circle (45);
\item inside of the figure (3, 10, 15, 17, 46);
\item outside of the figure (4, 11, 12, 18);
\item either inside or outside of the figure (38);
\item either inside or outside or on a side of the figure (32);
\item in a certain part of the plane of the figure (28).
\end{enumerate}
\item parallel to the base of the proposed figure (5, 14, 19, 21--23, 29, 33, 35, 54);
\item parallel to a diameter of the circle (49, 50).
\end{enumerate}
\item It is required to draw more than one transversal
(\textit{a}) through one point (9, 47, 48, 56); (\textit{b}) through
two points (13); (\textit{c}) parallel to one another, the
problem is indeterminate (51).
\item It is required to draw a circle (52).
\end{enumerate}
% -----File: 030.png---Folio 18------------------------------------------------------
\item Auxiliary Propositions:

Although not explicitly stated or proved, Leonardo makes
use of four out of six of Euclid's auxiliary propositions\footnotemark[113].
On the other hand he proves two other propositions which Favaro
does not number: (1) Triangles with one angle of the one
equal to one angle of the other, are to one another as the
rectangle formed by the sides about the one angle is to that
formed by the sides about the equal angle in the other;
(2) the medians of a triangle meet in a point and trisect one
another.
\end{enumerate}



% -----File: 031.png---Folio 19---------------------------------------------------

\Chapter{II}                                                              % Chapter II
\thispagestyle{empty}  % this may be redundant

\Paragraph{JORDANUS NEMORARIUS}{18}                                       % Paragraph 18
\phantomsection\pdfbookmark[0]{Abraham Savasorda, Jordanus Nemorarius,
                               Luca Paciuolo}{hyperPar18}
\textit{Abraham Savasorda, Jordanus Nemorarius, Luca
Paciuolo.}---In earlier articles (10, 11) incidental reference was
made to Leonardo's general indebtedness to previous writers
in preparing his \textit{Practica Geometriae}, and also to the debt
which later writers owe to Leonardo. Among the former,
perhaps mention should be made of Abraham bar Chijja ha
Nasi\Footnote{53}{                                                        % Footnote 53
  That is, Abraham son of Chijja the prince.
  \textit{Cf.}\ \textsc{Steinschneider},
  \textit{Bibliotheca Mathematica}, 1896, (2),
  \textsc{x}, 34--38, and \textsc{Cantor},
  \textit{Vorlesungen über Geschichte d.~Math.}
  \textsc{i}$_3$, 797--800, 907.}                                         % end footnote 53
of Savasorda and his \textit{Liber embadorum} known through
the Latin translation of Plato of Tivoli. Abraham was a
learned Jew of Barcelona who probably employed Plato of
Tivoli to make the translation of his work from the Hebrew.
This translation, completed in 1116, was published by Curtze,
from fifteenth century MSS., in 1902\Footnote{54}{                        % Footnote 54
  \textsc{M.~Curtze},
  ``Urkunden zur Geschichte der Mathematik im Mittelalter und
  der Renaissance\ldots'' Erster Teil
  (\textit{Abhandlung zur Geschichte der Mathematischen
  Wis\-sen\-schaf\-ten\ldots} XII.\ Heft), Leipzig, 1902, pp.~3--183.}.   % end footnote 54
Pages 130--159 of this
edition contain ``capitulum tertium in arearum divisionum
explanatione'' with Latin and German text, and among the
many other propositions given by Savasorda is that of
Proclus-Euclid (= Woepcke 28 = Leonardo 57). Compared
with Leonardo's treatment of divisions Savasorda's seems
rather trivial. But however great Leonardo's obligations to
other writers, his originality and power sufficed to make
a comprehensive and unified treatise.

Almost contemporary with Leonardo was Jordanus Nemorarius
(d.~1237) who was the author of several works, all
probably written before 1222.
% -----File: 032.png---- Folio 20 -----------------------------------------------------
Among these is \textit{Geometria vel De Triangulis}\Footnote{55}{         % Footnote 55
  Edited with Introduction by \textsc{Max Curtze},
  \textit{Mitteilungen des Copernicus--Vereins
  für Wis\-sen\-schaf\-ten und Kunst zu Thorn}.
  \textsc{vi}.\ Heft, 1887. In his discussion
  of the second book, \textsc{Cantor}
  (\textit{Vorlesungen ü.\ Gesch.\ d.\ Math.} \textsc{ii}$_2$, 75)
  is misleading and inaccurate.
  One phase of his inaccuracy has been referred to by \textsc{Eneström}
  (\textit{Bibliotheca Mathematica},
  Januar, 1912, (3), \textsc{xii}, 62).}                                  % end footnote 55
in four books. The second book is principally
devoted to problems on divisions: Propositions 1--7 to the
division of lines and Propositions 8, 13, 17, 18, 19 to the
division of rectilineal figures. The enunciations of Propositions
8, 13, 17, 19 correspond, respectively, to Euclid 3, 26, 19, 14
and to Leonardo 2, 4, 3, 36. But Jordanus's proofs are quite
differently stated from those of Euclid or Leonardo. Both
for themselves and for comparison with the Euclidean proofs
which have come down to us, it will be interesting to reproduce
propositions 13 and 17 of Jordanus.

``13. \textit{Triangulo dato et puncto extra ipsum signato
lineam per punctum transeuntem designare, que triangulum
per equalia parciatur}'' [pp.~15--16].

\CenteredGraphic{80mm}{032}                                               % Graphic in Paragraph 18

``Let $\mathit{abc}$ be the triangle and $d$ the point outside but
contained within the lines $\mathit{aef, hbl}$, which are lines dividing
the triangle equally and produced. For if $d$ be taken in any
such place, draw $\mathit{dg}$ parallel to $\mathit{ca}$ meeting $\mathit{cb}$ produced in $g$.
Join $\mathit{cd}$ and find $\mathit{mn}$ such that
\[ \triangle\mathit{cdg}:\triangle\mathit{aec}\;
  (= \tfrac{1}{2}\, \triangle \mathit{abc})=\mathit{cg:mn}. \]
% -----File: 033.png---Folio 21------------------------------------------------------
Then divide $\mathit{cg}$ in $k$ such that
\[\mathit{gk:kc}=\mathit{kc:mn}.   \]
Produce $\mathit{dk}$ to meet $\mathit{ca}$ in $p$. Then I say that $\mathit{dp}$ divides the
triangle $\mathit{abc}$ into equal parts.

For, since the triangle $\mathit{ckp}$ is similar to the triangle $\mathit{kdg}$,
by 4 of sixth\Footnote{56}{                                               % Footnote 56
  That is, Euclid's \textit{Elements}, \textsc{vi}.~4.}                   % end footnote 56
and parallel lines and 15 of first and definitions
of similar areas,
\[
\triangle\mathit{ckp}:\triangle\mathit{kdg}=\mathit{mn:kg}
\]
by corollary to 17 of sixth\Footnote{57}{                                 % Footnote 57
  I do not know the MS.\ of Euclid here referred to; but manifestly it is the
  Porism of \textit{Elements} \textsc{vi}.~19 which is quoted:
  ``If three straight lines be proportional,
  then as the first is to the third, so is the figure described on the first to that
  which is similar and similarly described on the second.''}.             % end footnote 57
But
\[
\triangle\mathit{kdg}:\triangle\mathit{cdg}=\mathit{kg:cg}.
\]

Therefore, by equal proportions,
\begin{DPalign*}
            \triangle\mathit{ckp}&:\triangle\mathit{cdg}=\mathit{mn:cg}.  \\
 \therefore \triangle\mathit{ckp}&:\triangle\mathit{cdg}=
   \triangle\mathit{aec}:\triangle\mathit{cdg}.  \\
\lintertext{And} \triangle\mathit{ckp}&=\triangle\mathit{aec}\; (=
    \tfrac{1}{2}\triangle\mathit{abc})
\end{DPalign*}
by 9 of fifth, and this is the proposition.

\noindent
And by the same process of deduction we may be led to
an absurdity, namely, that all may equal a part if the point $k$
be otherwise than between $e$ and $b$ or the point $p$ be otherwise
than between $h$ and $a$; the part cut off must always be either
all or part of the triangle $\mathit{aec}$.''

``17. \textit{Puncto infra propositum trigonum dato lineam per
ipsum deducere, que triangulum secet per equalia}'' [pp.~17--18].

\CenteredGraphic{80mm}{034}                                               % Graphic in Paragraph 18

``Let $\mathit{abc}$ be the triangle and $d$ the point inside and
contained within the part between $\mathit{ag}$ and $\mathit{be}$ which divide
two sides and triangle into equal parts. Through $d$ draw $\mathit{fdh}$
parallel to $\mathit{ac}$ and draw $\mathit{db}$.
Then by 12 of this book\Footnote{58}{                                     % Footnote 58
  That is, \textit{De Triangulis}, Book~2, Prop.~12:
  ``Data recta linea aliam rectam
  inuenire, ad quam se habeat prior sicut quilibet datus triangulus ad quemlibet
  datum triangulum'' [p.~15].}                                            % end footnote 58
draw $\mathit{mn}$ such that
\[
\mathit{bf:mn}=\triangle\mathit{bdf}:\triangle\mathit{bec}\;(= \tfrac{1}{2}\triangle\mathit{abc}).
\]
% -----File: 034.png---Folio 22---------------------------------------------------

\begin{DPalign*}
 \lintertext{Also find $\mathit{ty}$ such that} & \\
 \mathit{bf:ty} &= \triangle\mathit{bfh}:\triangle\mathit{bec}.  \\
 \lintertext{And since}  \triangle\mathit{bfh} >&
    \triangle\mathit{bdf},\quad \mathit{mn} > \mathit{ty}
\end{DPalign*}
by 8 and 10 of fifth.
\begin{DPalign*}
&\lintertext{Now}  \mathit{bf:bc} = \mathit{bc:ty}
\end{DPalign*}
by corollary to 17 of sixth\Footnote{581}{                                % Footnote 581 (was 58a)
  Rather is it the converse of this corollary, which is quoted in note~57. It
  follows at once, however:
\[
\mathit{bf:ty}=\triangle\mathit{bfh}:\triangle\mathit{bec}=
 \mathit{bf}^2:\mathit{bc}^2,\quad
 \therefore \mathit{bf}\centerdot\mathit{ty} =
     \mathit{bc}^2 \text{ or }\mathit{bf:bc}=\mathit{bc:ty}.
 \]},\DPnote{the comma follows the footnote in main text}                 % end footnote 581
and $\triangle\mathit{bfh} < \triangle\mathit{bec}$ since $\mathit{fh, ce}$
are parallel lines.



\begin{DPalign*}
\lintertext{\indent But} \mathit{bc:ty} > \mathit{bc:mn}
\end{DPalign*}
by second part of 8 of fifth.
\begin{gather*}
 \therefore \mathit{bf:bc} > \mathit{bc:mn}; \\
 \therefore \mathit{fc} < \tfrac{1}{4} \mathit{mn}
\end{gather*}
by 6 of this book\Footnote{59}{                                           % Footnote 59
  ``Cum sit linee breuiori adiecte major proporcio ad compositam, quam
  composite ad longiorem, breuiorem quarta longioris minorem esse necesse est
  [p.~13].}.\\                                                            % end footnote 59
% -----File: 035.png---Folio 23------------------------------------------------------
Add then to the line $\mathit{cf}$, from $f$, a line $\mathit{fz}$,
by 5 of this book\Footnote{60}{                                           % Footnote 60
  ``Duabus lineis propositis, quarum una sit minor quarta alterius uel equalis,
  minori talem lineam adiungere, ut, que adiecte ad compositam, eadem sit composite
  ad reliquam propositarum proporcio'' [p.~12].},                         % end footnote 60
such that
\[ \mathit{fz:zc} = \mathit{zc:mn}; \]
and $\mathit{fz}$ will be less than $\mathit{fb}$ by the first part of the premise.
[Supposition with regard to $d$?]

Join $\mathit{zd}$ and produce it to meet $\mathit{ac}$ in $k$; then I say that
the line $\mathit{zdk}$ divides the triangle $\mathit{abc}$ into equal parts. For
\[
 \triangle\mathit{bdf} : \triangle\mathit{zdf}=\mathit{bf:zf}
\]
by 1 of sixth.
\begin{DPalign*}
\lintertext{\indent But} \triangle\mathit{zdf} : \triangle\mathit{zkc}=\mathit{zf:mn}
\end{DPalign*}
by corollary to 17 of sixth\footnotemark[57] and similar triangles.\\
\indent Therefore by 1 and by equal proportions
\begin{DPalign*}
 \triangle\mathit{bdf} &: \triangle\mathit{zkc}=\mathit{bf:mn}.\\
\lintertext{\indent But} \triangle\mathit{bdf} &: \triangle\mathit{bec}=\mathit{bf:mn}.
\end{DPalign*}
\indent Therefore by the second part of 9 of fifth
\begin{DPalign*}
\triangle\mathit{zkc} &= \triangle\mathit{bec} = \tfrac{1}{2} \triangle\mathit{abc}."  \tag*{\textsc{q.e.f.}\quad}
\end{DPalign*}

Proposition 18 of Jordanus is devoted to finding the centre
of gravity of a triangle\Footnote{601}{                                   % Footnote 601 (was 60a)
  Archimedes proved (\textit{Works of Archimedes}, Heath ed., 1897, p.~201;
  \textit{Opera omnia} iterum edidit J.~L.\ Heiberg,
  \textsc{ii}, 150--159, 1913) in Propositions 13--14, Book~\textsc{i}
  of ``On the Equilibrium of Planes'' that
  \textit{the centre of gravity of any triangle is at
  the intersection of the lines drawn from any two angles
  to the middle points of the
  opposite sides respectively}.}                                          % end footnote 601
and it is stated in the form of a problem
on divisions. In Leonardo this problem is treated\footnotemark[109]
by showing that the medians of a triangle are concurrent; but
in Jordanus (as in Heron\footnotemark[83])
the question discussed is, ``to find a
point in a triangle such that when it is joined to the angular
points, the triangle will be divided into three equal parts''(p.~18).

A much later work, \textit{Summa de Arithmetica Geometria
Proportioni et Proportionalita\ldots} by Luca Paciuolo (b.~about
1445) was published at Venice in 1494\Footnote{61}{                       % Footnote 61
  A new edition appeared at Toscolano in 1523,
  and in the section which we are
  discussing there does not appear to be any material change.}.           % end footnote 61
In the geometrical
section (the second, and separately paged) of the work,
pages 35 \textit{verso}--43 \textit{verso}, problems on divisions of figures are
solved, and in this connection the author acknowledges great
debt to Leonardo's work. Although the treatment is not as
% -----File: 036.png---Folio 24------------------------------------------------------
full as Leonardo's, yet practically the same figures are employed.
The Proclus-Euclid propositions which have to do
with the division of a circle are to be found here.

\Paragraph{ARABIAN WRITERS ON DIVISIONS OF FIGURES}{19}                   % Paragraph 19
\phantomsection\pdfbookmark[0]{Arabian writers on Divisions of Figures}{hyperPar19}
\textit{``Muhammed Bagdedinus'' and other Arabian writers
on Divisions of Figures.}---We have not considered so far
who ``Muhammed Bagdedinus'' was, other than to quote the
statement of Dee\footnotemark[35]
that he may have been ``that \textit{Albategnus}
whom Copernicus often cites as a very considerable author, or
that \textit{Machomet} who is said to have been Alkindus's scholar.''
Albategnius or Muhammed b.\ Gâbir b.\ Sinân, Abû `Abdallâh,
el Battânî who received his name from Battân, in Syria,
where he was born, lived in the latter part of the ninth and in
the early part of the tenth century\Footnote{62}{                         % Footnote 62
  \textsc{M.~Cantor},
  \textit{Vorlesungen ü.\ Gesch.\ d.~Math}. \textsc{i}$_{3}$, 736.}.      % end footnote 62
El-Kindî (d. about 873) the
philosopher of the Arabians was in his prime about 850\Footnote{63}{      % Footnote 63
  \textsc{M.~Cantor},
  \textit{Vorlesungen ü.~Gesch.\ d.~Math}. \textsc{i}$_{3}$, 718.}.       % end footnote 63
``Alkindus's scholar'' would therefore possibly be a contemporary
of Albategnius. It is probably because of these
suggestions of Dee\Footnote{64}{                                          % Footnote 64
  \textit{Cf}.\ \textsc{Steinschneider}\footnotemark[12].}                % end footnote 64
that Chasles speaks\Footnote{65}{                                         % Footnote 65
  \textsc{Chasles}, \textit{Aper\c{c}u historique\ldots}
  $3^{e}$ éd., Paris, 1889, p.~497.}                                      % end footnote 65
of ``Mahomet
Bagdadin, géomètre du $\textrm{x}^e$ siècle.''

It would be scarcely profitable to do more than give
references to the recorded opinions of other writers such
as Smith\Footnote{66}{                                                    % Footnote 66
  \textsc{T.~Smith}, \textit{Vitae quorumdam\ldots virorum},
1707, p.~56. \textit{Cf}.\ notes 14, 15.},                                % end footnote 66
Kästner\Footnote{67}{                                                     % Footnote 67
  \textsc{A.~G.~Kästner},
  \textit{Geschichte der Mathematik\ldots},
  Band \textsc{i}, Göttingen, 1796, p.~273.
  See also his preface to \textsc{N.~Morville},
  \textit{Lehre von der geometrischen und ökonomischen
  Vertheilung der Felder, nach der dänischen Schrift bearbeitet
  von J.~W.~Christiani, begleitet mit einer Vorrede\ldots
  von A.~G.\ Kästner}, Göttingen, 1793.},                                 % end footnote 67
Fabricius\Footnote{68}{                                                   % Footnote 68
  \textsc{J.~A.\ Fabricius},
  \textit{Bibliotheca Graeca\ldots } Editio nova. Volumen quartum,
  Hamburgi, \textsc{mdcclxxxv}, p.~81.},                                  % end footnote 68
Heilbronner\Footnote{69}{                                                 % Footnote 69
  \textsc{J.~C.\ Heilbronner},
  \textit{Historia Matheseos universae\ldots}
  Lipsiae, \textsc{mdccxlii}, p.~438, 163--4.},                           % end footnote 69
Montucla\Footnote{70}{                                                    % Footnote 70
  \textsc{J.~F.\ Montucla},
  \textit{Histoire des mathématiques\ldots}\
  éd.\ nouv. Tome \textsc{i}, An \textsc{vii}, p.~216.},                  % end footnote 70
Hankel\Footnote{71}{                                                      % Footnote 71
  \textsc{H.~Hankel},
  \textit{Zur Geschichte der Math.\ in Alterthum u.\ Mittelalter},
  Leipzig, 1874, p.~234.},                                                % end footnote 71
Grunert\Footnote{72}{                                                     % Footnote 72
  \textsc{J.~A.~Grunert},
  \textit{Math. Wörterbuch \ldots von G.S. Klügel, fortgesetzt
  von C.~B.\ Mollweide und beendigt von J.A. Grunert\ldots}Erste
  Abteilung, die reine Math.,
  fünfter Theil, erster Band, Leipzig, 1831, p. 76.}--- whose             % end footnote 72
results Favaro summarizes\Footnote{73}{                                   % Footnote 73
  \textsc{Favaro}, pp.~141--144.}.                                         % end footnote 73

The latest and most trustworthy research in this connection
seems to be due to Suter who first surmised\Footnote{74}{                 % Footnote 74
  \textsc{H.~Suter},
  ``Die Mathematiker und Astronomen der Araber und ihre Werke''
  (\textit{Abh.\ z.\ Gesch.\ d.\ Math.\ Wiss.}
  \textsc{x}. Heft, Leipzig, 1900), p.~202, No.~517.}                     % end footnote 74
that the author
% -----File: 037.png---Folio 25------------------------------------------------------
of the Dee book On Divisions was
Mu\d{h}.\ b.\ Mu\d{h}.\ el-Ba\.{g}dâdî
who wrote at Cairo a table of sines for every minute. A little
later\Footnote{75}{                                                       % Footnote 75
  \textsc{H.~Suter}, \textit{idem}, ``Nachträge und Berichtigung''
  (\textit{Abh.\ z.\ Gesch.\ d.\ Math.\
  Wiss.}\ \textsc{xiv}.\ Heft, 1902), p.~181;
  also \textit{Bibliotheca Mathematica},
  \textsc{iv}$_3$, 1903, pp.~22--27.},                                    % end footnote 75
however, Suter discovered facts which led him to
believe that the true author was
Abû Mu\d{h}ammed b.\ `Abdelbâqî el-Ba\.{g}dâdî
(d.~1141 at the age of over 70 years) to whom
an excellent commentary on Book \textsc{x} of the \textit{Elements}
has been ascribed. Of a MS.\ by this author Gherard of Cremona
(1114--1187) may well have been a translator.

Euclid's book \textit{On Divisions} was undoubtedly the ultimate
basis of all Arabian works on the same subject. We have
record of two or three other treatises.

1. \b{T}âbit b.\ Qorra (826--901) translated parts of the works
of Archimedes and Apollonius, revised Ishâq's translation of
Euclid's \textit{Elements} and \textit{Data} and also revised the work
\textit{On Divisions of Figures} translated by an anonymous
writer\Footnote{76}{                                                      % Footnote 76
  \textsc{H.~Suter}, ``Die Mathematiker\ldots,'' pp.~34--38.}.            % end footnote 76

2. Abû Mu\d{h}.\ el-Hasan b.\ `Obeidallâh b.\ Soleimân b.\
Wahb (d.~901) was a distinguished geometer who wrote
``A Commentary on the difficult parts of the work of Euclid''
and ``The Book on Proportion.'' Suter thinks\Footnote{77}{                % Footnote 77
  \textsc{H.~Suter},
  ``Die Mathematiker\ldots,'' pp.\ 48 and 211, note 23.}                  % end footnote 77
that another
reading is possible in connection with the second title, and
that it may refer to Euclid's work \textit{On Divisions}.

3. Abû'l Wefâ el-Bûz\v{g}ânî (940--997) one of the greatest
of Arabian mathematicians and astronomers spent his later
life in Bagdad, and is the author of a course of Lectures on
geometrical constructions. Chapters \textsc{vii--ix} of the Persian
form of this treatise which has come down to us in roundabout
fashion were entitled: ``On the division of triangles,'' ``On
the division of quadrilaterals,'' ``On the division of circles''
respectively. Chapter \textsc{vii} and the beginning of Chapter \textsc{viii}
are, however, missing from the Bibliothèque nationale Persian
MS.\ which has been described by Woepcke\Footnote{78}{                    % Footnote 78
  \textsc{F.~Woepcke},
  ``Recherches sur l'histoire des Sciences mathématiques chez
  les orientaux, d'après des traités inedits Arabes et Persans.
  Deuxième article.
  Analyse et extrait d'un recueil de constructions géométriques
  par Aboûl Wafâ,''
  \textit{Journal asiatique},
  Fevrier--Avril, 1855, (5), V, 218--256, 309--359; reprint, Paris,
  1855, pp.~89.}.                                                         % end footnote 78
This MS.,
which gives constructions without demonstrations, was made
from an Arabian text, by one Abû Ishâq b.\ `Abdallâh with
% -----File: 038.png---Folio 26------------------------------------------------------
the assistance of four pupils and the aid of another translation.
The Arabian text was an abridgment of Abû'l Wefâ's lectures
prepared by a gifted disciple.

The three propositions of Chapter \textsc{ix}\Footnote{79}{               % Footnote 79
  \textsc{F.~Woepcke},
  \textit{idem}, pp.~340--341; reprint, pp.~70--71.}                      % end footnote 79
are practically
identical with Euclid (Woepcke) 28, 29.
In Chapter \textsc{viii}\Footnote{80}{                                    % Footnote 80
  \textsc{F.~Woepcke},
  \textit{idem}, pp.~338--340; reprint, pp.~68--70.}                      % end footnote 80
there are 24 propositions. About a score are given, in substance,
by both Leonardo and Euclid.

In conclusion, it may be remarked that in Chapter \textsc{xii}
of Abû'l Wefâ's work are 9 propositions, with various solutions,
for dividing the surface of a sphere into equiangular and
equilateral triangles, quadrilaterals, pentagons and hexagons.

\Paragraph{HERON OF ALEXANDRIA, APOLLONIUS OF PERGA}{20}                  % Paragraph 20
\phantomsection\pdfbookmark[0]{Practical applications}{hyperPar20}

\textit{Practical applications of the problems
On Divisions of Figures;
the \ItalicGreekSnippet{metrik'a}
of Heron of Alexandria}.---The popularity
of the problems of Euclid's book \textit{On Divisions} among Arabians,
as well as later in Europe, was no doubt largely due to the
possible practical application of the problems in the division
of parcels of land of various shapes, the areas of which,
according to the Rhind papyrus, were already discussed in
empirical fashion about 1800 \textsc{b.~c.} In the first century before
Christ\Footnote{81}{                                                      % Footnote 81
  This date is uncertain,
  but recent research appears to place it not earlier
  than 50~\textsc{b.~c.}\ nor later than 150~\textsc{a.~d.}
  \textit{Cf.}~\textsc{Heath},
  \textit{Thirteen Books of Euclid's Elements},
  \textsc{i}, 20--21; or perhaps better still, Article
  ``Heron~5'' by K.~Tittel in Pauly-Wissowa's
  \textit{Real-Encyclopädie der class. Al\-tert\-ums\-wis\-sen\-schaf\-ten},
  \textsc{viii}, Stuttgart,
  1913, especially columns 996--1000.}                                    % end footnote 81
we find that Heron of Alexandria dealt with the
division of surfaces and solids in the third book of his
\textit{Surveying} (\ItalicGreekSnippet{metrik'a})\Footnote{82}{          % Footnote 82
  \textit{Heronis Alexandrini opera quae supersunt omnia},
  Vol.~\textsc{iii}, \textit{Rationes Dimetiendi et commentatio
  Dioptrica recensuit Hermannus Schoene}, Lipsiae,
  \textsc{mcmiii}. Third
  book, pp.~140--185. \textit{Cf.}\ \textsc{Cantor},
  \textit{Vorlesungen\ldots}, \textsc{i}$_3$, 380--382.}.                 % end footnote 82
Although the enunciations of the
propositions in this book are, as a whole, similar\Footnote{83}{          % Footnote 83
  Only two are exactly the same:
  \textsc{ii--iii} (= Euclid~30), \textsc{vii} (= Euclid~32),
  the problem considered in \textsc{x} is practically Euclid~27 (Art.~48),
  while \textsc{xviii} is closely
  related to Euclid~29 (Art.~50).
  In \textsc{xix} Heron finds in a triangle a point such that
  when it is joined to the angular points, the triangle will be
  divided into three equal parts. The divisions of solids of which
  Heron treats are of a sphere (\textsc{xxiii}) and the division in a
  given ratio, by a plane parallel to the base, of a Pyramid
  (\textsc{xx}) and of a Cone (\textsc{xxi}).
  For proof of Proposition \textsc{xxiii}:
  \textit{To cut a sphere by a plane so that the volumes of the segments
  are to one another in a given ratio}, Heron refers to Proposition~4,
  Book~\textsc{ii} of ``On the Sphere and Cylinder'' of Archimedes; the
  third proposition in the same book of the Archimedean work is
  (Heron~\textsc{xvii}):
  \textit{To cut a given sphere by a plane so that the surfaces
  of the segments may have to one another a given ratio}.
  (\textit{Works of Archimedes}, Heath ed., 1897, pp.~61--65;
  \textit{Opera omnia}
  iterum edidit J.~L.\ Heiberg, \textsc{i}, 184--195, 1910.)

  Propositions \textsc{ii} and \textsc{vii} are also given in Heron's
%[DPStyle Greek: peri dioptras]
  \ItalicGreekSnippet{per`i di'optras}
 (Schoene's edition, pp.~278--281).
 \textit{Cf}.\ ``Extraits des Manuscrits relatifs à la géométrie grecs''
% typo corrected: single quote should be double.
  par A.~J.~C.\ Vincent,
  \textit{Notices et extraits des Manuscrits de la bibliothèque impériale},
  Paris, 1858, \textsc{xix}, pp.\ 157, 283, 285.}                         % end footnote 83
to those
% -----File: 039.png---Folio 27------------------------------------------------------
in Euclid's book \textit{On Divisions}, Heron's discussion consists
almost entirely of ``analyses'' and approximations. For
example, \textsc{ii}: ``To divide a triangle in a given ratio by a
line drawn parallel to the base''---while Euclid gives the
general construction, Heron considers that the sides of the
given triangle have certain known numerical lengths and
thence finds the approximate
distance of the angular points of
the triangle to the points in the sides where the
required line
parallel to the base intersects them, because, as he expressly
states, in a field with uneven surface it is difficult to draw a
line parallel to another.
Most of the problems are discussed
with a variety of numbers although theoretical analysis sometimes
enters.
Take as an example Proposition \textsc{x}\Footnote{84}{                   % Footnote 84
  \textsc{Heron}, \textit{idem}, p.~160f.}:                               % end footnote 84
``\textit{To divide
a triangle in a given ratio by a line drawn from a point in a
side produced.}''

\WrappedGraphic{60mm}{039}{}{r}                                           % Graphic in Paragraph 20

``Suppose the construction made. Then the ratio of triangle
$\mathit{AEZ}$ to quadrilateral $\mathit{ZEB\Gamma}$ is known;
also the ratio of the triangle $\mathit{AB\Gamma}$ to
the triangle $\mathit{AZE}$. But the triangle
$\mathit{AB\Gamma}$ is known, therefore so is the
triangle $\mathit{AZE}$. Now $\Delta$ is given.
Through a known point $\Delta$ there is
therefore drawn a line which, with two
lines $\mathit{AB}$ and $\mathit{A\Gamma}$ intersecting in $A$,
encloses a known area.

Therefore the points $E$ and $Z$ are
given. This is shown in the second
book of \textit{On Cutting off a Space}. Hence the required proof.

If the point $\Delta$ be not on $\mathit{B\Gamma}$ but anywhere this will make
no difference.''

\Paragraph{HERON OF ALEXANDRIA, APOLLONIUS OF PERGA}{21}                  % Paragraph 21
\phantomsection\pdfbookmark[0]{Apollonius of Perga}{hyperPar21}

\textit{Connection between Euclid's book On Divisions,
Apollonius's treatise On Cutting off a Space and a Pappus-lemma
to Euclid's book of Porisms}.---Although the name of
the author of the above-mentioned work is not given by
Heron, the reference is clearly to Apollonius's lost work.
According to Pappus it consisted of two books which contained
124 propositions treating of the various cases of the
% -----File: 040.png---Folio 28------------------------------------------------------
following problem: \textit{Given two coplanar straight lines
$\mathit{A_1P_1,B_2P_2}$, on which $A_1$ and $B_2$ are fixed points; it is required to
draw through a fixed point $\Delta$ of the plane, a transversal $\mathit{\Delta ZE}$
forming on $\mathit{A_1P_1, B_2P_2}$ the two segments $\mathit{A_1Z, B_2E}$ such that
$\mathit{A_1Z}\centerdot\mathit{B_2E}$ is equal to a given rectangle.}

Given a construction for the particular case when
$\mathit{A_1P_1,B_2P_2}$ meet in $A$,
and when $A_1$ and $B_2$ coincide with $A$---Heron's
reasoning becomes clear. The solution of this particular
case is practically equivalent to the solution of Euclid's
Proposition 19 or 20 or 26 or 27. References to restorations
of Apollonius's work are given in note 111.

To complete the list of references to writers before 1500,
who have treated
\WrappedGraphic{70mm}{040}{11}{r}                                         % Graphic in Paragraph 21
of Euclid's
problems here under discussion,
I should not fail to mention the
last of the 38 lemmas which
Pappus gives as useful in connection
with the 171 theorems
of Euclid's lost book of
\textit{Porisms:
Through a given point $E$ in $\mathit{BD}$
produced to draw a line cutting
the parallelogram $\mathit{AD}$ such that
the triangle $\mathit{Z\Gamma H}$ is equal to the parallelogram $\mathit{AD}$.}

After ``Analysis'' Pappus has the following

``Synthesis. Given the parallelogram $\mathit{AD}$ and the point $E$. Through $E$
draw the line $\mathit{EZ}$ such that the
rectangle $\mathit{\Gamma Z\centerdot \Gamma H}$ equals twice the
rectangle $\mathit{A\Gamma\centerdot \Gamma D}$. Then
according to the above analysis [which contains a reference
to an earlier lemma discussed a little later\footnotemark[88]
in this book] the
triangle $\mathit{Z\Gamma H}$ equals the parallelogram $\mathit{AD}$. Hence $\mathit{EZ}$
satisfies the problem and is the only line to do so\Footnote{85}{         % Footnote 85
  Pappus ed.\ by Hultsch, Vol.~2, Berlin, 1877, pp.~917--919.
  In Chasles's restoration
  of Euclid's \textit{Porisms},
  this lemma is used in connection with ``Porism \textsc{clxxx}:
  Given two lines $\mathit{SA,SA'}$, a point $P$ and a space $\nu$:
  points $I$ and $J'$ can be found in
  a line with $P$ and such that if one take on $\mathit{SA}$,
  $\mathit{SA'}$ two points $m$, $m'$, bound by the
  equation $\mathit{Im}\centerdot\mathit{J'm'}= \nu$,
  the line $\mathit{mm'}$ will pass through a given point.''
  \textit{Les trois livres de Porismes d'Euclide},
  Paris, 1860, p.~284. See also the restoration by
  R.~Simson, pp.~527--530 of ``De porismatibus tractatus,''
  \textit{Opera quaedam reliqua\ldots}
  Glasguae, \textsc{m.dcc.lxxvi.}}.''                                     % end footnote 85

The tacit assumption here made, that the equivalent of
a proposition of Euclid's book \textit{On Divisions (of Figures)} was
well known, is noteworthy.


% -----File: 041.png---Folio 29------------------------------------------------------

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\Chapter{III}                                                             % Chapter III
%% Setting this chapter as \raggedbottom alleviates many of the
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\raggedbottom
\phantomsection\pdfbookmark[0]{Restoration of Euclid's MS.}{hyperCh3}

\bigskip

\LargeCenteredNote{``\textit{The Treatise of Euclid on the Division} (\textit{of plane Figures}).''}

\thispagestyle{empty}
%%%%%%\newpage

% -----File: 042.png---Folio 30------------------------------------------------------

\bigskip\bigskip\bigskip

\Proposition{1}                                                           % Proposition 1
\Paragraph{}{22}                                                          % Paragraph 22
\textit{``To divide\Footnote{86}{                                         % Footnote 86
  Literally, the original runs, according to Woepcke,
  ``We propose to ourselves to demonstrate how to divide, etc.''
  I have added all footnotes except those attributed to Woepcke.}         % end footnote 86
a given triangle into equal parts by
a line parallel to its base.''}
[Leonardo 5, p.~119, ll.~7--9.]

Let $\mathit{abg}$ be the given triangle which it is required to bisect by a line
parallel to $\mathit{bg}$. Produce $\mathit{ba}$ to
$d$ till $\mathit{ba = 2ad}$. Then in $\mathit{ba}$ find a point
$e$ such that
\[
\mathit{ba:ae} = \mathit{ae:ad}.
\]
Through $e$ draw $\mathit{ez}$ parallel to $\mathit{bg}$; then
the triangle $\mathit{abg}$ is divided by the line
$\mathit{ez}$ into two equal parts, of which one
is the triangle $\mathit{aez}$, and the other the
quadrilateral $\mathit{ebgz}$.

\WrappedGraphic{45mm}{042}{}{r}                                           % Graphic in Proposition 1

\SmallLeftNote{Leonardo then gives three proofs, but as the
first and second are practically equivalent, I shall
only indicate the second and third.}

I. When three lines are proportional, as the first is to the
third so is a figure on the first to the similar and similarly
situated figure described on the second [\textsc{vi}.~19, ``Porism'']\Footnote{87}{%
Throughout the restoration I have added                                   % Footnote 87
  occasional references of this kind to Heath's edition of Euclid's
  \textit{Elements}; \textsc{vi}.~19 refers to Proposition~19 of
  Book~\textsc{vi}.  \textit{Cf.}\ note~57.}.                             % end footnote 87
\[
\therefore\mathit{ba:ad}=\text{figure on }\mathit{ba}: \text{similar and similarly situated
figure on }\mathit{ae}.
\]
\begin{DPalign*}
\lintertext{\indent Hence} \mathit{ba:ad} &= \triangle\mathit{abg} : \triangle\mathit{aez}\\
                                          &= 2:1.
\end{DPalign*}
\[
 \therefore \triangle  \mathit{abg} = 2 \triangle\mathit{aez}.
\]
\begin{DPalign*}
\lintertext{\indent II.}  \mathit{ba:\;}  \mathit{ae} &= \mathit{ae:ad}. \\
\therefore \mathit{ba} & \centerdot\mathit{ad}=\mathit{ae}^2,
\end{DPalign*}
% -----File: 043.png---Folio 31------------------------------------------------------
and since $\mathit{ad}$ is one-half of $\mathit{ba}$,
\[
ba^2 = 2\,ae^2.
\]
And since $\mathit{bg}$ is parallel to $\mathit{ez}$,
\begin{flalign*}
&& ba : ae &= ga : az.      &&\\
&& \therefore ba^2 : ae^2 &= ga^2 : az^2. \qquad &&\text{ [\textsc{vi}.~22]} \quad\\
&\text{\indent But } & ba^2 &= 2\,ae^2. &&\\
&& \therefore ga^2 &= 2\,az^2. &&\\
&\text{\indent Then }
& ba\centerdot ag &= 2\,ae\centerdot az, &&\text{ [\textsc{vi}.~22]} \quad\\
&& \therefore \triangle\mathit{abg} &= 2\,\triangle\mathit{aez}\,\footnotemark[88].  &&
\end{flalign*}

\SmallLeftNote{Then follows a numerical example.}

\Footnotetext{88}{The theorem here assumed is enunciated by               % Footnote 88 (long)
  Leonardo (p.~111, ll.~24--27) as follows:
%% The graphic is part of the footnote. There are several of these.
\WrappedGraphic{49mm}{043}{}{r}                                           % Graphic in Proposition 1
  \textit{Et si á trigono
  recta protracta fuerit secans duo latera trigonj, que cum ipsis
  duobus lateribus faciant trigonum habentem angulum unum comunem cum
  ipso trigono, erit proportio unius trigoni ad alium, sicut facta ex
  lateribus continentibus ipsum angulum}. This is followed by the
  sentence ``Ad cuius rei euidentiam.'' Then come the construction and
  proof:

  Let $\mathit{abc}$ be the given triangle and $\mathit{de}$ the line
  across it, meeting the sides $\mathit{ca}$ and $\mathit{cb}$ in the points
  $d$, $e$, respectively. I say that
  \[
    \triangle\mathit{abc}:\triangle\mathit{dec} =
       \mathit{ac}\centerdot\mathit{cb:dc}\centerdot\mathit{ce}.
  \]

  \textit{Proof:}
  To $\mathit{ac}$ apply the triangle $\mathit{afc}= \triangle\mathit{dec}$. [\textsc{i}.~44]

  Since the triangles $\mathit{abc, afc}$ are of the same altitude,
  \begin{flalign*}
    && \mathit{bc} &: \mathit{fc} = \triangle\mathit{abc} : \triangle\mathit{afc}.
    &&\text{ [\textsc{vi}.~1]\quad}\\
    &\text{\indent But}
    & \mathit{bc} &: \mathit{fc} = \mathit{ac}\centerdot\mathit{bc}:\mathit{ac}\centerdot\mathit{fc},
    &&\text{ [\textsc{v}.~15]\quad}  \\
    && \therefore \triangle\mathit{abc} &: \triangle\mathit{afc} =
    \mathit{ac}\centerdot\mathit{bc} : \mathit{ac}\centerdot\mathit{fc},\qquad\qquad&&\\
    &\text{and since } \triangle\mathit{dec} = \triangle\mathit{acf},&&&&\\
    && \triangle\mathit{acb} &: \triangle\mathit{dce} =
        \mathit{ac}\centerdot\mathit{bc}:\mathit{ac}\centerdot\mathit{cf}. &&
  \end{flalign*}

  Again, since the triangles $\mathit{acf, dce}$ are equal and have a common angle, as in
  the fifteenth theorem of the sixth book of Euclid, the sides are mutually proportional.
  \begin{gather*}
    \therefore \mathit{ac:dc}=\mathit{ce:cf},\qquad
    \therefore \mathit{ac}\centerdot\mathit{cf}=\mathit{dc}\centerdot\mathit{ce},\\
    \therefore \triangle\mathit{acb}:\triangle\mathit{dce} =
                \mathit{ac}\centerdot\mathit{cb}:\mathit{dc}\centerdot\mathit{ce}.\\
  \end{gather*}
  \hfill ``quod oportebat ostendere.''\qquad

  It is to be observed that the Latin letters are used with the above figure.
  This suggests the possibility of the proof being due to Leonardo.

  The theorem is assumed in Euclid's proof of proposition~19 (Art.~40)
  and it occurs, directly or indirectly, in more than one of his
  works. A proof, depending on the proposition that the area of a
  triangle is equal to one-half the product of its base and altitude,
  is given by Pappus (pp.~894--897) in connection with one of his
  lemmas for Euclid's book of \textit{Porisms}: \textit{Triangles
  which have one angle of the one equal or supplementary to one angle
  of the other are in the ratio compounded of the
% -----File: 044.png---Folio 32------------------------------------------------------
  ratios of the sides about the equal or supplementary angles}.
  (\textit{Cf.}\ \textsc{R.~Simson},
  ``De Porismatibus Tractatus'' in
  \textit{Opera quaedam reliqua\ldots} 1776,
  p.~515 ff.---\textsc{P.~Breton} (de Champ),
  ``Recherches nouvelles sur les porismes d'Euclide,''
  \textit{Journal de mathématiques pures et appliquées},
  \textsc{xx}, 1855, p.~233 ff. Reprint, p.~25 ff.---\textsc{M.~Chasles},
  \textit{Les trois livres de Porismes d'Euclide\ldots}
  Paris, 1860, pp.\ 247, 295, 307.)

  The first part of this lemma is practically equivalent to either
  (1) [\textsc{vi}.~23]:
  \textit{Equiangular parallelograms have to one another the ratio
  compounded of the ratio of their sides};
  or (2) the first part of Prop.~70 of the \textit{Data}
  (\textit{Euclidis Data\ldots} edidit H.~Menge, Lipsiae, 1896, p.~130f.):
  \textit{If in two equiangular parallelograms the sides
  containing the equal angles have a given ratio to one another}
  [i.~e.\ one side in one to one side in the other],
  \textit{the parallelograms themselves will also have a given ratio to
  one another.} \textit{Cf.}\ \textsc{Heath}, \textit{Thirteen Books of
  Euclid's Elements}, \textsc{ii}, 250.

  The proposition is stated in another way by Pappus\footnotemark[85]
  (p.~928) who proves that
  \textit{a parallelogram is to an equiangular parallelogram as the
  rectangle contained by the adjacent sides of the first is to the
  rectangle contained by the adjacent sides of the second.}

  The above theorem of Leonardo is precisely the first of those
  theorems which Commandinus adds to \textsc{vi}.~17 of \emph{his}
  edition of Euclid's \textit{Elements} and concerning which he writes
  ``à nobis elaborata'' (``fatti da noi''):
  \textit{Euclidis Elementorum Libri XV\ldots
  A Federico Commandino\ldots}Pisauri,
  \textsc{mdlxxii}, p.~81
  \textit{recto (Degli Elementi d' Euclide libri quindici con gli
  scholii antichi tradotti prima in lingua latina da }
  M.~Federico Commandino
  \textit{da Urbino, et con commentarii illustrati, et hora d' ordine
  dell' istesso transportati nella nostra vulgare, et da lui riveduti.}
  In Urbino, \textsc{m.d.lxxv}, p.~88
  \textit{recto}). }                                                      % end footnote 88 (long)

\Proposition{2}                                                           % Proposition 2
\Paragraph{}{23} \textit{``To divide a given triangle into three equal    % Paragraph 23
parts by two lines parallel to its base.''}
[Leonardo 14, p.~122, l.~8.]

\WrappedGraphic{45mm}{044}{}{r}                                           % Graphic in Proposition 2

Let $\mathit{abg}$ be the given triangle with base $\mathit{bg}$.
Produce $\mathit{ba}$
to $d$ till $\mathit{ba} = 3\mathit{ad}$, and produce $\mathit{ad}$ to
$e$ till $\mathit{ad}=\mathit{de}$; then $ae = \frac{2}{3} ba$. Find
$\mathit{az}$, a mean proportional between $\mathit{ba}$
and $\mathit{ad}$, and $\mathit{ia}$ a mean proportional
between $\mathit{ba}$ and $\mathit{ae}$. Then through
$z$ and $i$ draw $\mathit{zt}$, $\mathit{ik}$ parallel to $\mathit{bg}$
and I say that the triangle $\mathit{abg}$ is
divided into three equal parts of
which one is the triangle $\mathit{azt}$, another
the quadrilateral $\mathit{zikt}$, the third
the quadrilateral $\mathit{ibgk}$.

\textit{Proof:} Since
\begin{gather*}
  \mathit{ba:az} = \mathit{az:ad},  \\
  \mathit{ba:ad} = \triangle\mathit{abg}:\triangle\mathit{azt}, \tag*{[\textsc{vi}.~19, Porism]}
\end{gather*}
for these triangles are similar.
% -----File: 045.png---Folio 33------------------------------------------------------
\begin{DPalign*}
\lintertext{\indent Now }  ba = 3ad; \quad\therefore &\triangle\mathit{abg} =
   3\triangle\mathit{azt}.  \\
  \therefore\triangle\mathit{azt} &= \tfrac{1}{3}\triangle\mathit{abg}. &&  \\
\lintertext{\indent Again,}  ba:ia &= ia:ae;
\end{DPalign*}
$\therefore ba: ae = \triangle$ on $\mathit{ea}$: similar and similarly situated $\triangle$ on $\mathit{ai}$.

But triangles $\mathit{aik, abg}$ are similar and similarly described
on $\mathit{ai}$ and $\mathit{ab}$; and
\begin{align*}
\mathit{ea:ab} &= 2: 3.  \\
  \therefore\triangle\mathit{aik} &= \tfrac{2}{3}\triangle\mathit{abg}.
\end{align*}

And since $\triangle\mathit{azt} = \tfrac{1}{3}\triangle\mathit{abg}$, there remains the quadrilateral
$\mathit{zikt} = \tfrac{1}{3} \triangle\mathit{abg}$.
We see that the quadrilateral $\mathit{ibgk}$ will be the
other third part; hence the triangle $\mathit{abg}$ has been divided into
three equal parts; ``quod oportebat facere.''

\SmallLeftNote{Leonardo continues:}
``Et sic per demonstratos modos omnia
genera trigonorum possunt diuidi in quatuor partes uel
plures.''
\SmallLeftNote{Cf.~note 45.}

\Proposition{3}                                                           % Proposition 3
\Paragraph{}{24} \textit{``To divide a given triangle into two equal      % Paragraph 24
parts by a line drawn from a given
point situated on one of the sides
of the triangle.''}
[Leonardo 1, 2,
p.~110, l.~31; p.~111, ll.~41--43.]

\WrappedGraphic{45mm}{045}{}{r}                                           % Graphic in Proposition 3

Given the triangle $\mathit{bgd}$; if $a$
be the middle point of $\mathit{gd}$ the line
$\mathit{ba}$ will divide the triangle as required;
either because the triangles
are on equal bases and of
the same altitude [\textsc{i}.~38; Leonardo
1], or because
\begin{flalign*}
  && \triangle\mathit{bgd}: \triangle\mathit{bad}
&= \mathit{bd}\centerdot\mathit{dg}: \mathit{bd}\centerdot\mathit{da}\footnotemark[88].\qquad  &&\\
\text{\indent Whence } && \triangle\mathit{bgd} &= 2\triangle\mathit{bad}. &&
\end{flalign*}

% -----File: 046.png---Folio 34------------------------------------------------------

But if the given point be not the middle point of any side,
let $\mathit{abg}$ be the triangle and
$d$ the given point nearer to $b$
than to $g$. Bisect $\mathit{bg}$ at $e$ and
draw $\mathit{ad,ae}$.
Through $e$ draw $\mathit{ez}$ parallel to $\mathit{da}$; join $\mathit{dz}$.
Then the triangle $\mathit{abg}$ is bisected by $\mathit{dz}$.

\WrappedGraphic{60mm}{046}{7}{r}                                          % Graphic in Proposition 3

\textit{Proof:} Since
\[
\mathit{ad}\parallel\mathit{ez},\quad \triangle\mathit{adz} = \triangle\mathit{ade}.
\]

To each add $\triangle\mathit{abd}$. Then
\begin{DPalign*}
 \text{quadl.}\mathit{abdz}      &=
    \triangle\mathit{abd} + \triangle\mathit{ade},  \\
                                  &= \triangle\mathit{abe}.    \\
 \lintertext{\indent But}  \triangle\mathit{abe} &= \tfrac{1}{2} \triangle\mathit{abg}; \\
 \therefore\, \text{quadl.}\mathit{abdz} &= \tfrac{1}{2} \triangle\mathit{abg};
\end{DPalign*}
and the triangle $\mathit{zdg}$ is the other half of the triangle $\mathit{abg}$.
Therefore the triangle $\mathit{abg}$ is divided into two equal parts by
the line $\mathit{dz}$ drawn from the point $d$;

\hfill ``ut oportebat facere.''\qquad\qquad

\SmallCenteredNote{Then follows a numerical example.}

\Proposition{4}                                                           % Proposition 4
\Paragraph{}{25} \textit{``To divide a given                              % Paragraph 25
  trapezium\Footnote{89}{%                                                % Footnote 89
  Here, and in what follows, this word is used to refer to a
  quadrilateral two of whose sides are parallel.}                         % end footnote 89
into two equal parts by a line parallel to its base.''}
[Leonardo~23, p.~125, ll.~37--38.]

Let $\mathit{abgd}$ be the given trapezium with parallel sides
$\mathit{ad,bg,ad}$ being the lesser. It is required to bisect the trapezium
by a line parallel to the base $\mathit{bg}$.
Let $\mathit{gd,ba}$, produced, meet in a point $e$.
Determine $z$ such that
\[
ze^2=\tfrac{1}{2}(eb^2+ea^2).\footnotemark[90]                            % Footnotemark 90
\]
\Footnotetext{90}{                                                        % Footnotetext 90
  The point $z$ is easily found by constructions which twice make
  use of \textsc{i}.~47.}                                                 % end of footnote 90
Through $z$ draw $\mathit{zi}$ parallel to $\mathit{gb}$.
I say that the trapezium $\mathit{abgd}$ is divided into two equal
parts by the line $\mathit{zi}$ parallel to the base $\mathit{bg}$.

% -----File: 047.png---Folio 35------------------------------------------------------

\WrappedGraphic{60mm}{047}{16}{r}                                         % Graphic in Proposition 4

\textit{Proof:} For since
\[
2ze^2 = eb^2 + ea^2,
\]
and all the triangles are similar,
\[
2\triangle\mathit{ezi} = \triangle\mathit{ebg} + \triangle\mathit{ead}.
                     \tag*{\text{[\textsc{vi}. 19]}}
\]

From the triangle $\mathit{ebg}$ take
away the triangle $\mathit{ezi}$. Then
\[
\triangle\mathit{ezi} = \text{quadl.}\mathit{zbgi} + \triangle\mathit{eda}.
\]

And taking away from the
equals the triangle $\mathit{eda}$, we get
\[
\text{quadl.}\mathit{ai} = \text{quadl.}\mathit{zg}.
\]

Therefore the trapezium
$\mathit{abgd}$ is divided into two equal
parts by the line $\mathit{zi}$ parallel to its base. \qquad\textsc{q.~o.~f.}

\SmallCenteredNote{A numerical example then follows.}

\Proposition{5}                                                           % Proposition 5
\Paragraph{}{26} \textit{``And we divide the given trapezium into three   % Paragraph 26
equal parts as we divide the triangle, by a construction analogous to
the preceding construction\footnotemark[91].''}
[Leonardo~33, p.~134, ll.~14--15.]

Let $\mathit{abgd}$ be the trapezium with parallel sides $\mathit{ad,bg}$ and
other sides $\mathit{ba, gd}$ produced to meet in $e$. Let $\mathit{zti}$ be a line
such that
\[
\mathit{zi:it} = eb^2:ea^2.\footnotemark[92]
\]

\Footnotetext{91}{                                                        % Footnote 91
  It is to be noticed that Leonardo's discussion of this proposition
  is hardly ``analogous to the preceding construction'' which is
  certainly simpler than if it had been similar to that of Prop.~5.
  A construction for Prop.~4 along the same lines,
  which may well have been Euclid's method, would obviously be as follows:

  Let $\mathit{zti}$ be a line such that
  \[
    \mathit{zi:it} = eb^2:ea^2.
  \]

  Divide $\mathit{tz}$ into two equal parts, $\mathit{tk,kz}$. Find $m$ such that
  \[
    em^2:eb^2 = ki:zi.
  \]

  Then $m$ leads to the same solution as before. [For, in brief,
  \begin{align*}
    em^2 &= eb^2\left(\frac{ki}{zi}\right)
      = eb^2\left(\frac{\frac{zt}{2} + ti}{zi}\right)
      = \frac{eb^2}{2}\left(\frac{zi + it}{zi}\right)
      = \frac{eb^2}{2}\left(\frac{ea^2 + eb^2}{eb^2}\right)\\
      &= \tfrac{1}{2}\left(ea^2 + eb^2\right).\text{]}
  \end{align*}}                                                           % end footnote 91

\Footnotetext{92}{                                                        % Footnote 92
  From \textsc{vi.}~19, Porism, it is clear that the construction
  here is to find a line $x$ which is a third proportional to $\mathit{eb}$
  and $\mathit{ea}$. Then $zi:it = eb:x$.}                                % end footnote 92

% -----File: 048.png---Folio 36------------------------------------------------------

Divide $\mathit{tz}$ into three equal parts $\mathit{tk, kl, lz}$.

Find $m$ and $n$ in $\mathit{be}$ such that
\begin{DPalign*}
                     em^2 : eb^2 &= ik : zi,  \\
  \lintertext{and}   en^2 : eb^2 &= il : zi.
\end{DPalign*}

Through $m$ and $n$ draw $\mathit{mo,np}$ parallel to the base $\mathit{bg}$.
Then I say that the quadrilateral $\mathit{ag}$ is divided into three
equal parts: $\mathit{ao,mp,ng}$.
\begin{DPalign*}
  \lintertext{\indent \textit{Proof:} For}
  eb^2 : ea^2 & = \triangle\mathit{ebg} : \triangle\mathit{ead}.
  \tag*{[\text{\textsc{vi.}~19}]\quad}  \\
  \therefore zi : it &= \triangle\mathit{ebg} : \triangle\mathit{ead}.
  \tag*{\ldots\ldots\ldots\ldots\ldots[1]\quad}
\end{DPalign*}
\CenteredGraphic{60mm}{048}                                               % Graphic in Proposition 5
\begin{DPalign*}
  \lintertext{\indent But } \mathit{zi:ik} &= eb^2 : em^2, \\
  \therefore \mathit{zi:ik} &= \triangle\mathit{ebg} : \triangle\mathit{emo}.
  \tag*{\dots[2]\quad}\\
  \lintertext{\indent So also}
  \mathit{zi:il} &= \triangle\mathit{ebg} : \triangle\mathit{enp}.
  \tag*{\dots[3]\quad}\\
  \lintertext{Whence}
  \mathit{it:tk} &= \triangle\mathit{ead} : \text{quadl.}\mathit{ao},\footnotemark[93] \\
  \lintertext{and therefore}
  \mathit{tk:kl} &= \text{quadl.}\mathit{ao} : \text{quadl.}\mathit{mp}.\footnotemark[94]
\end{DPalign*}
\Footnotetext{93}{                                                        % Footnote 93
  This may be obtained by combining [1] and [2],
  and applying \textsc{v.}\ 11, 16, 17.}                                  % end footnote 93
\Footnotetext{94}{                                                        % Footnote 94
  Relations [1], [2] and [3] may be employed, as in the preceding,
  to give,
  \[
  \mathit{it:kl} = \triangle\mathit{ead} : \text{quadl.}\mathit{mp};
  \]
  combining this with $\mathit{it:tk} = \triangle\mathit{ead} : \text{quadl.}\mathit{ao}$,
  we get the required result,
  \[
  \mathit{tk:kl} = \text{quadl.}\mathit{ao} : \text{quadl.}\mathit{mp}.
  \]}                                                                     % end footnote 94
% -----File: 049.png---Folio 37------------------------------------------------------
\begin{DPgather*}
  \lintertext{\indent But} \mathit{tk} = \mathit{kl}. \quad \therefore \text{quadl.}\mathit{ao}
    = \text{quadl.}\mathit{mp}.   \\
  \lintertext{So also} \mathit{kl:lz} = \text{quadl.}\mathit{mp}: \text{ quadl.}\mathit{ng}; \\
  \lintertext{and} \mathit{kl} = \mathit{lz}. \quad \therefore \text{quadl.}\mathit{mp}
    = \text{quadl.}\mathit{ng}.
\end{DPgather*}

Therefore the quadrilateral is divided into equal quadrilaterals
$\mathit{ao, mp, ng}$; \hfill ``ut prediximus.''

\SmallCenteredNote{Then follows a numerical example.}

\Proposition{6}                                                           % Proposition 6
\Paragraph{}{27} \textit{``To divide a parallelogram into two equal parts % Paragraph 27
by a straight line drawn from a given point situated on one of the
sides of the parallelogram.''}
[Leonardo 16, p.~123, ll.~30--31.]

Let $\mathit{abcd}$ be the parallelogram and $i$ any point in the side $\mathit{ad}$.
Bisect $\mathit{ad}$
\WrappedGraphic{70mm}{049}{}{r}                                           % Graphic in Proposition 6
in $f$ and $\mathit{bc}$ in $e$.
Join $\mathit{fe}$. Then the parallelogram
$\mathit{ac}$ is divided into equal
parallelograms $\mathit{ae}$, $\mathit{fc}$ on equal bases.

Cut off $\mathit{eh=fi}$.  Join $\mathit{hi}$.
Then this is the line required.

\SmallLeftNote{Leonardo gives two proofs:}

I. Let $\mathit{hi}$ meet $\mathit{fe}$ in $k$.
Then [$\triangle$s $\mathit{fki, hke}$ are equal;
add to each the pentagon $\mathit{kfabh}$, etc.]

II. Since $\mathit{ae, fc}$ are \Rect \,s,
$\mathit{af=be}$ and $\mathit{fd=ec}$. But
\begin{gather*}               %%\PGnote this could just be a gather
  \mathit{fd}         = \tfrac{1}{2} \mathit{ad}.   \\
  \therefore \mathit{fd} = \mathit{af} = \mathit{ec}.
\end{gather*}
\begin{DPgather*}
  \lintertext{\indent And since } \mathit{fi} = \mathit{he},  \quad \mathit{ai} = \mathit{ch}.  \\
  \lintertext{\indent So also }
    di = bh, \text{ and } hi \text{ is common.}   \\
  \therefore \text{quadl.}\mathit{iabh} = \text{quadl.}\mathit{ihcd}.\footnotemark[95]
\end{DPgather*}

\Footnotetext{95}{                                                        % Footnote 95
  The first rather than the second proof is Euclidean.
  There is no proposition of the \textit{Elements} with regard to the
  equality of quadrilaterals whose sides and angles, taken in the same
  order, are equal. Of course the result is readily deduced from
  \textsc{i}.~4, if we make certain suppositions with regard to order.
  Cf.\ the proof of Prop.~10.}                                            % end of footnote 95

% -----File: 050.png---Folio 38------------------------------------------------------

Similarly if the given point were between $a$ and $f$, [etc.;
or on any other side]. And thus a parallelogram can be
divided into two equal parts by a straight line drawn from a
given point situated on any one of its sides.

\Proposition{7}                                                           % Proposition 7
\Paragraph{}{28} \textit{``To cut off a certain fraction from a given     % Paragraph 28
parallelogram by a straight line drawn from a given point situated
on one of the sides of the parallelogram.''}
[Leonardo 20 (the case where the fraction is one-third), p.~124, ll.~24--26.]

Let $\mathit{abcd}$ be the given parallelogram. Suppose it be
required to cut off a third of this parallelogram, by a straight
line drawn from $i$, in the side $\mathit{ad}$.

\CenteredGraphic{70mm}{050}                                               % Graphic in Proposition 7

\SmallCenteredNote{(The figure here is a combination of two in the original.)}

Trisect $\mathit{ad}$ in $e$ and $f$ and through $e$, $f$ draw $\mathit{eg, fh}$ parallel
to $\mathit{dc}$;
[then these lines trisect the \Rect\,.
If the point $i$ be in
the line $\mathit{ad}$, at either $e$ or $f$, then the problem is solved. But
if it be between $a$ and $e$,
draw $\mathit{ik}$ to bisect the \Rect \,$\mathit{ah}$
(Prop.~6), etc. Similarly if $i$ were between $e$ and $f$, or
between $f$ and $d$].

\SmallLeftNote{After finishing these cases Leonardo concludes:}

``eodem modo potest \DPtypo{omnem}{omne}
paralilogramum diuidi in
quatuor uel plures partes equales\footnotemark[45].''

\small
The construction in this proposition is limited to the case where ``a certain
fraction'' is the reciprocal of an integer. But more generally, if the fraction
were $\mathit{m:n}$ (the ratio of the lengths of two given lines),
we could proceed in a
very similar way: Divide $\mathit{ad}$ in $e$, internally,
so that $\mathit{ae : ed = m:n-m \quad (n > m)}$.
In $\mathit{ad}$ cut off $\mathit{ef}=\mathit{ae}$ and
through $f$ draw $\mathit{fh}$ parallel to $\mathit{ab}$.
Then, as before, the problem is reduced to Proposition~6.

If the point $e$ should fall at $i$ or in the interval $\mathit{ai}$ the
part cut off from the
parallelogram by the required line would be in the form of a triangle which
might be determined by \textsc{i}.~44.
\normalsize

% -----File: 051.png---Folio 39------------------------------------------------------

\Proposition{8}                                                           % Proposition 8
\Paragraph{}{29} \textit{``To divide a given trapezium into two equal     % Paragraph 29
parts by a straight line drawn from a given point situated on the
longer of the sides of the trapezium.''}
[Part of Leonardo~27\Footnote{96}{                                        % Footnote 96
  Leonardo~27: ``Quomodo quadrilatera duorum laterum equidistantium
  diuidantur á puncto dato super quodlibet latus ipsius''
  [p.~129, ll.~2--3]. \textit{Cf}.\ note 46.},                            % end footnote 96
p.~127, ll.~2--3.]

\SmallLeftNote{
This enunciation means, apparently, ``from a given point situated on the
longer of the [parallel] sides.'' At any rate Leonardo gives constructions for
the cases when the given point is on any side. These I shall take up
successively. The figure is made from more than one of Leonardo's, and
there is a slight change in the lettering.}

\CenteredGraphic{110mm}{051}                                              % Graphic in Proposition 8

Let $\mathit{ad}$ be the shorter of the parallel sides $\mathit{ad, bg}$, which
are bisected in $t$ and $k$ respectively. Join $\mathit{tk}$. Then if $\mathit{bt,gt}$
be joined, [it is clear, from triangles on equal bases and
between the same parallels, that $\mathit{tk}$ bisects the trapezium].
[This is Leonardo~24, p.~126, l.~31.]

\SmallLeftNote{Next consider the given point as any point on the shorter side}
[Leonardo~25, p.~127, ll.~2--3].

First let the point be at the angle $a$. Cut off $\mathit{kl}$ in $\mathit{kg}$,
equal to $\mathit{at}$.
Join $\mathit{al}$, meeting $\mathit{tk}$ in $m$; then the quadrilateral
is divided as required by $\mathit{al}$. For [the triangles $\mathit{atm, mkl}$ are
equal in all respects, etc.].

Similarly if $d$ were the given point; in $\mathit{kb}$ cut off $\mathit{kn}$ equal
to $\mathit{td}$, and $\mathit{dn}$ divides the quadrilateral into two equal parts
which is proved as in the preceding case.

% -----File: 052.png---Folio 40------------------------------------------------------

[Were the given point anywhere between $a$ and $t$ the
other end of the bisecting line would be between $k$ and $l$.
Similarly if the given point were between $t$ and $d$, the corresponding
point would be between $k$ and $n$.]

\SmallLeftNote{Although not observed by Favaro, Leonardo now considers:}

If the given point be in the side $\mathit{bg}$; either $l$, or $n$, or a
point between $l$ and $n$, then the above construction is at once
applicable.

Suppose, however, that the given point were at $b$ or in
the segment $\mathit{bn}$, at $g$ or in the segment $\mathit{lg}$.
First consider the given point at $b$.
Join $\mathit{bd}$ and through $n$ draw $\mathit{nc}$ parallel
to $\mathit{bd}$ to meet $\mathit{gd}$ in $c$. Join $\mathit{bc}$.
Then $\mathit{bc}$ bisects the trapezium.
For [$\mathit{abnd}$ is half of the trapezium $\mathit{ag}$, and the triangle $\mathit{bnd}$
equals the triangle $\mathit{bdc}$ etc.].

Similarly from a given point between $b$ and $n$, a line could
be drawn meeting $\mathit{gd}$ between $c$ and $d$, and dividing the
quadrilateral into two equal parts.

So also from $g$ a line $\mathit{gf}$ could be drawn [etc.]; and
similarly for a given point between $g$ and $l$.

\SmallLeftNote{Leonardo then concludes (p.~127, ll.~37--40):}

``Jam ostensum est quomodo in duo equa quadrilatera
duorum equidistantium laterum diuidi debeant á linea protracta
ab omni dato puncto super lineas equidistantes ipsius;
nunc uero ostendamus quomodo diuidantur á linea egrediente
á dato puncto super reliqua latera.''

\SmallLeftNote{
This is overlooked by Favaro, though implied in his 27 [Leonardo, p.~129, l.~4].
I may add Leonardo's discussion of the above proposition although it
does not seem to be called for by Euclid.}

\WrappedGraphic{60mm}{052}{}{r}                                           % Graphic in Proposition 8

Let the point be in the side $\mathit{gd}$. For $g$ or $c$ or $d$ or any
point between $c$ and $d$ the above
constructions clearly suffice. Let
us, then, now consider the given
point $h$ as between $c$ and $g$.
Draw the line $\mathit{iz}$ parallel to $\mathit{gb}$
to bisect the trapezium (Prop.~4).
Suppose $h$ were between $g$ and $i$.
Join $\mathit{zh}$. Through $i$ draw $\mathit{ik}$
parallel to $\mathit{hz}$, and meeting $\mathit{ab}$
in $k$.

\SmallLeftNote{(The lettering of the original figure is somewhat changed.)}

% -----File: 053.png---Folio 41------------------------------------------------------
Join $\mathit{hk}$, then [this is the line required; since
\[
  \triangle\mathit{izh} = \triangle\mathit{kzh}, \text{ etc.]}.
\]

[Similarly if $h$ were between $i$ and $d$.]

[So also for points on the line $\mathit{ab}$.]

\Proposition{9}                                                           % Proposition 9
\Paragraph{}{30} \textit{``To cut off a certain fraction from a given     % Paragraph 30
trapezium by a straight line drawn from a given point situated on the
longer side of the trapezium.''}
[Leonardo 30, 31\Footnote{97}{                                            % Footnote 97
  As 30, Favaro quotes,
  ``Per rectam protractam super duo latera equidistantia
  quadrilaterum abscisum in data aliqua proportione dividere''; as 31:
  ``Divisionem in eadem proportione ab angulis habere.''},                % end of footnote 97
p.~133, ll.~17--19, 31.]

\small
I shall interpret ``longer side'' as in Proposition~8, and lead up to the
consideration of any given point on $\mathit{bg}$ after discussing the cases of points on
the shorter side $\mathit{ad}$.
\normalsize

\CenteredGraphic{95mm}{053}                                               % Graphic in Proposition 9

\SmallCenteredNote{(This figure is made from three of Leonardo's.)}

Suppose it be required to divide the trapezium in the ratio
$\mathit{ez:zi}\,$\Footnote{98}{                                          % Footnote 98
  Here, as well as in 15 and 36, Leonardo introduces the representation of
  numbers by straight lines, and in considering these lines he invariably writes the
  word number in connection with them;
  \textit{e.~g.}\ `number $\mathit{ez}$: number $\mathit{zi}$,' not $\mathit{ez:zi}$.
  Euclidean MSS.\ of the \textit{Elements},
  Books \textsc{vii} to \textsc{ix}, adopt this same method.
  In what follows, I shall use the abbreviated form.}.                    % end of footnote 98

Divide $\mathit{ad}$, $\mathit{bg}$ in the points $t$, $k$, respectively, such that
\[
  \mathit{at:td} = \mathit{ez:zi} = \mathit{bk:kg}.
\]
% -----File: 054.png---Folio 42------------------------------------------------------
Join $\mathit{tk}$.
Then by joining $\mathit{bt}$ and $\mathit{gt}$ [it is easily seen by \textsc{vi}.~1
and \textsc{v}.~12, that the trapezium $\mathit{ag}$ is
divided by $\mathit{tk}$ in the ratio $\mathit{ez:zi}$].

If the given point be at $a$ or $d$,
make $\mathit{kl}=\mathit{at}$ and $\mathit{gn}=\mathit{bl}$.
Join $\mathit{al,dn}$. [Adding the quadrilateral $\mathit{ak}$ to the congruent
triangles with equal sides $\mathit{at,kl}$, we find $\mathit{al}$ divides the
trapezium in the required ratio. Then from \textsc{vi}.~1, $\mathit{dn}$ does
the same.]

\SmallLeftNote{
As in Proposition 8, for any point $t'$ between $a$ and $t$, or $t$ and $d$, we have
a corresponding point $k'$ between $l$ and $k$ or $n$ and $k$, such that the line $t'k'$
divides the trapezium in the given ratio.

If the given point be in $\mathit{bg}$ at $l$ or $n$ or between $l$ and $n$, the above
reasoning suffices.}

Suppose however that the given point were at $b$. Join $\mathit{bd}$.
Through $n$ draw $\mathit{nc}$ parallel to $\mathit{bd}$.
Join $\mathit{bc}$. Then $\mathit{bc}$ divides
the trapezium in the required ratio. Similarly for the point $g$
and for any point between $b$ and $n$, or between $g$ and $l$.

\SmallLeftNote{Some of the parts which I have filled in above are covered by the general
final statement:}
``nec non et diuidemus ipsum quadrilaterum ab
omni puncto dato super aliquod laterum ipsius\ldots\ldots''
\SmallLeftNote{(Page 134, ll.~10--11. Compare Proposition 13.)}

\Proposition{10}                                                          % Proposition 10
\Paragraph{}{31} \textit{``To divide a parallelogram into two equal parts % Paragraph 31
by a straight line drawn from a given point outside the parallelogram.''}
[Leonardo 18, p.~124, ll.~5--7.]

\CenteredGraphic{90mm}{054}                                               % Graphic in Proposition 10

Let $\mathit{abcd}$ be the given parallelogram and $e$ the point outside.
Join $\mathit{bd}$ and bisect it in $g$.
Join $\mathit{eg}$ meeting $\mathit{bc}$ in $k$
% -----File: 055.png---Folio 43------------------------------------------------------
and produce it to meet $\mathit{ad}$ in $f$. Then the parallelogram has
been divided into two equal parts by the line drawn through $e$,
as may be proved by superposition; and one half is the quadrilateral
$\mathit{fabk}$,
the other, the quadrilateral $\mathit{fkcd}$\,\Footnote{99}{              % Footnote 99
  The proof also follows from the equality of the triangles
  $\mathit{fgd,bgk}$, by \textsc{i}.~26 and of the triangles
  $\mathit{abd,bdc}$ by
  \textsc{i}.~4. This problem is possible for all positions of the
  point $e$.}.                                                            % end footnote 99

\newpage
\Proposition{11}                                                          % Proposition 11
\Paragraph{}{32} \textit{``To cut off a certain fraction from a           % Paragraph 32
parallelogram by a straight line drawn from a given point
outside of the parallelogram.''}

\small
This proposition is not explicitly formulated by Leonardo; but the general
method he would have employed seems obvious from what has gone before.

Suppose it were required to cut off one-third of the given parallelogram $\mathit{ac}$
by a line drawn through a point $e$ outside of the parallelogram. Then by the
method of Proposition 7, form a parallelogram two-thirds of $\mathit{ac}$. There are four
such parallelograms with centres $g_1, g_2, g_3, g_4$. Lines $l_1, l_2, l_3, l_4$ through
each one of these points and $e$ will bisect a parallelogram (Proposition 10).

There are several cases to consider with regard to the position of $e$ but it
may be readily shown that, in one case at least, there is a line $l_i(i=1, 2, 3, 4)$,
which will cut off a third of the parallelogram $\mathit{ac}$.

Similarly for one-fourth, one-fifth, or any other fraction such as $\mathit{m:n}$ which
represents the ratio of lengths of given lines.
\normalsize

\Proposition{12}                                                          % Proposition 12
\Paragraph{}{33} \textit{``To divide a given trapezium into two equal     % Paragraph 33
parts by a straight line drawn from a point which is not situated
on the longer side of the trapezium. It is necessary that the point
be situated beyond the points of concourse of the two sides of the
trapezium.''}
[Leonardo 28, p.~129, ll.~2--4, and another, unnumbered\Footnote{100}{     % Footnote 100
  As Leonardo 28 Favaro gives,
``Qualiter quadrilatera duorum laterum
equidistantium \DPtypo{dividi}{diuidi}
debeant a dato puncto extra figuram''
and entirely ignores the paragraph headed,
``De diuisione eiusdem generis, qua quadrilaterorum
per rectam transeuntem per punctum datum infra ipsum''
[p.~131, ll.~13--14].}.]                                                   % end footnote 100

\Proposition{13}                                                          % Proposition 13
\Paragraph{}{34} \textit{``To cut off a certain fraction from a           % Paragraph 34
(parallel-) trapezium by a straight line which passes through a
given point lying inside or outside the trapezium but so that a
straight line can be drawn through it cutting both the parallel
% -----File: 056.png---Folio 44------------------------------------------------------
sides of the trapezium}\Footnote{101}{                                    % Footnote 101
  The final clauses of Propositions 12 and 13, in Woepcke's rendering,
  are the same. I have given a literal translation in Proposition~12.
  Heath's translation and interpretation (after Woepcke) are given in 13.
  Concerning 12 and 13 Woepcke adds the following note:
  ``Suppose it were required to cut off the $n$th part of the
  trapezium $\mathit{ABDC}$; make $\mathit{A\alpha}$ and
  $\mathit{C\gamma}$ respectively equal
  to the $n$th parts of $\mathit{AB}$ and of $\mathit{CD}$;
  then $\mathit{A\alpha\gamma C}$ will be the $n$th part of the trapezium,
  for $\mathit{\gamma\alpha}$ produced
  will pass through the intersection of $\mathit{CA}$, $\mathit{DB}$ produced.
  Now to draw through a given point $E$ the transversal which cuts off
  a certain fraction of the trapezium, join the middle point $\mu$ of
  the segment $\mathit{\alpha\gamma}$, and the point $E$, by a line;
  this line $\mathit{EFG}$ will be the transversal required to be drawn,
  since the triangle $\mathit{\alpha F\mu}$
  equals the triangle $\mathit{\gamma G\mu}$.

\CenteredGraphic{110mm}{056}                                              % Graphic in Proposition 13
% again, the graphic is part of the footnote!}

  ``But when the given point is situated as $E'$ or $E''$ such
  that the transversal drawn through $\mu$ no longer meets the two
  parallel sides but one of the parallel sides and one of the two
  other sides, or the other two sides; then the construction indicated
  is not valid since $\mathit{CG'\mu\gamma}$ is not equal to $\mathit{BF'\mu\alpha}$.
  It appears that this is the idea which the text is intended to express.
  The `points of concourse' are the vertices where a parallel side
  and one of the two other sides intersect; and the expression
  `beyond' refers to the movement of the transversal represented as
  turning about the point $\mu$.''}.''                                    % end footnote 101
[Part of Leonardo 32\Footnote{102}{                                       % Footnote 102
  ``Quadrilaterum [trapezium] ab omni puncto dato super aliquod laterum
  ipsius, et etiam ab omni puncto dato infra,
  uel extra diuidere in aliqua data proportioni.''}%                       % end footnote 102
, p.~134, ll.~11--12.]

\SmallLeftNote{We first take up Leonardo's discussion of Proposition 12.}

In the figure of Proposition 8, suppose $\mathit{al}$ to be produced
in the directions of the points $e$ and $r$;
$\mathit{tk}$ in the directions of $q$ and $v$,
$\mathit{dn}$ of $z$ and $h$,
$\mathit{cb}$ of $i$ and $o$,
$\mathit{gf}$ of $s$ and $p$.
Then for [any such exterior points
$e$, $q$, $z$, $i$, $s$, $r$, $v$, $h$, $o$, $p$,
lines are drawn bisecting the trapezium].

If the given point, $x$, were anywhere in the section of the
plane above $\mathit{ad}$ and between $\mathit{ea}$ and $\mathit{dz}$,
the line joining $x$ to $m$ would [by the same reasoning as in Proposition~8]
bisect the trapezium. Similarly for all points below $\mathit{nl}$ and between
% -----File: 057.png---Folio 45------------------------------------------------------
$\mathit{hn}$ and $\mathit{lr}$
[\ldots\ldots so also for all points within the triangles $\mathit{amd, nml}$].

\small
This seems to be all that Euclid's Proposition 12 calls for. But just as
Leonardo considers Proposition 8 for the general case with the given point
anywhere on the perimeter of the trapezium, so here, he discusses the constructions
for drawing a line from any point inside or outside of a trapezium
to divide it into two equal parts.

Leonardo does not give any details of the discussion of Euclid's Proposition
13, but after presentation of the cases given in Proposition 9
concludes:
\normalsize
``et diuidemus ipsum quadrilaterum ab omni puncto
dato super aliquod laterum ipsius, et etiam ab omni puncto
dato infra uel extra''
\small
[Leonardo 32, p.~134, ll.~10--12].

From Leonardo's discussion in Propositions 8, 9, 12, not only are the
necessary steps for the construction of 13 (indicated in the Woepcke note
above\footnotemark[101])
evident, but also those for the more general cases, not considered
by Euclid, where restrictions are not imposed on the position of the given
point.
\normalsize

\Proposition{14}                                                          % Proposition 14
\Paragraph{}{35} \textit{``To divide a given quadrilateral into two       % Paragraph 35
equal parts by a straight line drawn from a given vertex of the
quadrilateral.''} [Leonardo 36, p.~138, ll.~10--11.]

Let $\mathit{abcd}$ be the quadrilateral and $a$ the given vertex.
Draw the diagonal $\mathit{bd}$, meeting the diagonal $\mathit{ac}$ in $e$.
If $\mathit{be}$, $\mathit{ed}$
are equal, [$\mathit{ac}$ divides the quadrilateral as required].

If $\mathit{be}$ be not equal to $\mathit{ed}$,
make $\mathit{bz} = \mathit{zd}$.

Draw $\mathit{zi}\parallel\mathit{ac}$ to meet $\mathit{dc}$ in $i$.
Join $\mathit{ai}$. Then the quadrilateral
$\mathit{abcd}$ is divided as required by the line $\mathit{ai}$.

\CenteredGraphic{75mm}{057}                                               % Graphic in Proposition 14

\emph{Proof}: Join $\mathit{az}$ and $\mathit{zc}$. Then the triangles $\mathit{abz, azd}$ are
respectively equal to the triangles $\mathit{cbz, cdz}$.

% -----File: 058.png---Folio 46------------------------------------------------------

Therefore the quadrilateral $\mathit{abcz}$ is one-half the quadrilateral $\mathit{abcd}$.

And since the triangles $\mathit{azc,aic}$ are on the same base and
between the same parallels $\mathit{ac,zi}$, they are equal.

To each add the triangle $\mathit{abc}$.

Then the quadrilateral $\mathit{abcz}$ is equal to the quadrilateral
$\mathit{abci}$.
But the quadrilateral $\mathit{abcz}$ is one-half of the quadrilateral $\mathit{abcd}$.
Therefore $\mathit{abci}$ is one-half of the quadrilateral
$\mathit{abcd}$;      \hfill ``ut~oportet.''\qquad\qquad

\Proposition{15}                                                          % Proposition 15
\Paragraph{}{36} \textit{``To cut off a certain fraction from a given     % Paragraph 36
quadrilateral by a line drawn from a given vertex of the
quadrilateral.''} [Leonardo 40, p.~140, ll.~36--37.]

Let the given fraction be as $\mathit{ez:zi}$, and let the
quadrilateral be $\mathit{abcd}$ and
the given vertex $d$.
Divide $\mathit{ac}$ in $t$ such that
\[
  \mathit{at:tc} = \mathit{ez:zi}.
\]
If $\mathit{bd}$ pass through $t$ [then $\mathit{bd}$ is
the line required].

\WrappedGraphic{60mm}{058}{}{r}                                           % Graphic in Proposition 15

But if $\mathit{bd}$ do not pass through
$t$ it will intersect either $\mathit{ct}$ or $\mathit{ta}$;
let it intersect $\mathit{ct}$. Join $\mathit{bt}$, $\mathit{td}$.

Then
\[
  \text{quadl.}\mathit{tbcd} : \text{quadl.}\mathit{tbad}=\mathit{ct:ta}=\mathit{ez:zi}.
\]

Draw $\mathit{tl}$ parallel to the diagonal $\mathit{bd}$, and join $\mathit{dl}$.
Then the quadrilaterals $\mathit{lbcd, tbcd}$ are equal and the construction has
been made as required; for
\[
  \mathit{ct:ta}=\mathit{ez:zi} = \text{quadl.}\mathit{lbcd} : \triangle\mathit{dal}.
\]

And if $\mathit{bd}$ intersect $\mathit{ta}$ [a similar construction may be given
to divide the given quadrilateral, by a line through $d$, into a
quadrilateral and triangle in the required ratio].

\SmallLeftNote{Leonardo then gives the construction for dividing a quadrilateral in
a given ratio by a line drawn through a point which divides a side of the
quadrilateral in the given ratio.}

% -----File: 059.png---Folio 47------------------------------------------------------

\Proposition{16}                                                          % Proposition 16
\Paragraph{}{37} \textit{``To divide a given quadrilateral into two       % Paragraph 37
equal parts by a straight line drawn from a given point situated on
one of the sides of the quadrilateral.''}
[Leonardo 37, p.~138, ll.~28--29.]

Let $\mathit{abcd}$ be the given quadrilateral, $e$ the given point.
Divide $\mathit{ac}$ into two equal parts by the line $\mathit{dt}$ [Prop.~14].
Join $\mathit{et}$. The line $\mathit{et}$ either is, or is not,
parallel to $\mathit{dc}$.

\CenteredGraphic{90mm}{059}                                               % Graphic in Proposition 16

\SmallCenteredNote{(Two of Leonardo's figures are combined in one, here.)}

If $\mathit{et}$ be parallel to $\mathit{dc}$, join $\mathit{ec}$.
Then the quadrilateral $\mathit{ac}$ [is bisected by the line $\mathit{ec}$, etc.].

If $\mathit{et}$ be not parallel to $\mathit{dc}$,
draw $\mathit{dz} \parallel \mathit{et}$. Join $\mathit{ez}$.
Then $\mathit{ac}$ [is bisected by the line $\mathit{ez}$, etc.].

\SmallLeftNote{
Leonardo does not consider the case of failure of this construction, namely
when $\mathit{dz}$ falls outside the quadrilateral. Suppose in such a case that the
problem were solved by a line joining $e$ to a point $z'$ (not shown in the figure)
on $\mathit{dc}$. Through $t$, draw $\mathit{tt'} \parallel \mathit{cd}$.
Join $\mathit{ct'}$.
Then $\triangle\mathit{ct'd} = \triangle\mathit{ctd}= \triangle\mathit{edz'}$.
Whence $\triangle \mathit{et'c} = \triangle\mathit{ez'c}$,
or $\mathit{t'z'} \parallel \mathit{ce}$.
Therefore from $t'$, $z'$ may be found and the
solution in this case is also possible, indeed in more than one way, but it is
not in Euclid's manner to consider this question.}

Should the diagonal $\mathit{db}$ bisect the quadrilateral $\mathit{ac}$,
the discussion is similar to the above.

But if the line drawn from $d$ to bisect the quadrilateral
meet the side $\mathit{ab}$ in $i$,
draw $\mathit{bk}$ bisecting the quadrilateral $\mathit{ac}$.

If $k$ be not the given point, it will be between $k$ and $d$ or
between $k$ and $a$.
% -----File: 060.png---Folio 48------------------------------------------------------

In the first case join $\mathit{be}$ and through $k$ draw
$\mathit{kl}\parallel\mathit{eb}$.
Join $\mathit{el}$ [then $\mathit{el}$ is the required bisector].

\CenteredGraphic{120mm}{160}                                              % Graphic in Proposition 16

If the point $e$ be between $a$ and $k$ [a similar construction
with the line through $k$ parallel to $\mathit{be}$, and meeting $\mathit{bc}$ in $m$,
leads to the solution by the line $\mathit{em}$].

\CenteredGraphic{75mm}{260}                                               % Graphic in Proposition 16

Were $e$ at the middle of a side such as $\mathit{ab}$,
draw $\mathit{dz}\parallel\mathit{ab}$
and bisect $\mathit{dz}$ in $i$. Join $\mathit{ei}$, $\mathit{ci}$ and $\mathit{ec}$.
Through $i$ draw $\mathit{it}\parallel\mathit{ec}$.
Join $\mathit{et}$; then $\mathit{et}$ [bisects the quadrilateral
  $\mathit{ac}$,
since $\triangle\mathit{itc}=\triangle\mathit{ite}$,
etc.].

\SmallLeftNote{If $\mathit{dz}$ were to fall outside the quadrilateral,
draw from $c$ the parallel to $\mathit{ba}$;
and so on.}

% The page breaks generated by LaTeX often lead to unusual
%   blocks of white space. It was found experimentally that forcing
%   a page break at this point made the overall appearance much better.
%   I realize that this is a brittle fix.
%   Perhaps this will prove unneccessary in the future.}
\newpage
\Proposition{17}                                                          % Proposition 17
\Paragraph{}{38} \textit{``To cut off a certain fraction from a           % Paragraph 38
quadrilateral by a straight line drawn from a given point situated on
one of the sides of the quadrilateral.''}
[Leonardo~39, p.~140, ll.~11--12.]

% -----File: 061.png---Folio 49------------------------------------------------------

Let $\mathit{abcd}$ be the given quadrilateral and suppose it be
required to cut off one-third by a line drawn from the point $e$
in the side $\mathit{ad}$.

\CenteredGraphic{90mm}{161}                                               % Graphic in Proposition 17

Draw $\mathit{dz}$ cutting off one-third of $\mathit{ac}$ [Prop.~15].

Join $\mathit{ez}$, $\mathit{ec}$.

If $\mathit{ez}\parallel\mathit{dc}$, then $\mathit{ecd}$ [is the required part cut off, etc.].

But if $\mathit{ez}$ be not parallel to $\mathit{dc}$,
draw $\mathit{di}\parallel\mathit{ez}$ and join $\mathit{ei}$.
[Then this is the line required, etc.]

\SmallLeftNote{
The case when $\mathit{ei}$ cuts $\mathit{dc}$ is not taken up but it may be considered as in
the last proposition.

So also to divide $\mathit{ac}$ into any ratio: draw $\mathit{dz}$ dividing it in that ratio
(Prop.~15), and then proceed as above.

A particular case which Leonardo gives may be added.}

Let $\mathit{ab}$ be divided into three equal parts $\mathit{ae, ef, fb}$; draw
$\mathit{dg} \parallel \mathit{ab}$ and cut off $\mathit{gh} = \tfrac{1}{3}gd$.
Join $\mathit{fc}$ and through $h$ draw
$\mathit{hi} \parallel \mathit{fc}$, meeting $\mathit{dc}$ in $i$.
Join $\mathit{fi}$; and the quadrilateral $\mathit{fbci}$
will be one-third of the quadrilateral $\mathit{ac}$. [As in latter part of
Prop.~16.]

\CenteredGraphic{90mm}{261}  % originally 95mm                            % Graphic in Proposition 17

% -----File: 062.png---Folio 50------------------------------------------------------

Then $\mathit{ek}$ may be drawn to bisect the quadrilateral $\mathit{afid}$
[Prop.~16], and thus the quadrilateral $\mathit{abcd}$ will be divided
into three equal portions which are the quadrilaterals $\mathit{ak, ei, fc}$.


\Proposition{18}                                                          % Proposition 18
\nopagebreak
\Paragraph{}{39} \textit{``To apply to a straight line a rectangle equal  % Paragraph 39
to the rectangle contained by $\mathit{AB}$, $\mathit{AC}$ and deficient by a
square\footnotemark[103].''}

\Footnotetext{103}{                                                       % Footnote 103 (long)
  This proposition is interesting as illustrating the method of
  \emph{application of areas} which was
  ``one of the most powerful methods on which Greek Geometry relied.''
  The method first appears in the \textit{Elements} in \textsc{i}.~44:
  \textit{To a given straight line to apply, in a given rectilineal
  angle, a parallelogram equal to a given triangle}---a proposition
  which Heath characterises as
  ``one of the most impressive in all geometry''
  while the ``marvellous ingenuity of the solution is indeed worthy
  of the `godlike men of old' as Proclus calls the discoverers of the
  method of `application of areas'; and there would seem to be no
  reason to doubt that the particular solution, like the whole theory,
  was Pythagorean, and not a new solution due to Euclid himself.''

  [I continue to quote mainly from Heath who may be consulted for much
  greater detail: \textsc{Heath},
  \textit{Thirteen Books of Euclid's Elements},
  \textsc{i}, 9, 36, 343--7, 383--8;
  \textsc{ii}, 187, 257--67---\textsc{Heath},
  \textit{Apollonius of Perga Treatise on Conic Sections}, Cambridge,
  1896, pp.~lxxxi--lxxxiv, cii--cxi---\textsc{Heath},
  \textit{The Works of Archimedes}, Cambridge, 1897,
  pp.~xl--xlii, 110 and ``Equilibrium of Planes,''
  Bk~\textsc{ii}, Prop.~1, and ``On conoids and spheroids,''
  Props.\ 2, 25, 26, 29.
  See also: \textsc{Cantor},
  \textit{Vorlesungen über Geschichte der Math.}\ \textsc{i}$_3$, 289--291, etc.
  (under index heading `Flächenanlegung')---\textsc{H.~G.~Zeuthen},
  \textit{Geschichte der Mathematik im Alterthum und Mittelalter},
  Kopenhagen, 1896, pp.~45--52 (French ed.~Paris,
  1902, pp.~36--44)--- \textsc{C.~Taylor},
  \textit{Geometry of Conics}\ldots, Cambridge, 1881,
  pp.~\textsc{xliii--xliv}.]

  The simple \emph{application} of a parallelogram of given area to a
  given straight line as one of its sides is what we have in the
  \textit{Elements} \textsc{i}.~44 and 45; the general form
  of the problem with regard to \emph{exceeding} and
  \emph{falling-short} may be stated thus:

  ``To apply to a given straight line a rectangle
  (or, more generally, a parallelogram)
  equal to a given rectilineal figure and (1) \emph{exceeding}
  or (2) \emph{falling-short} by a square (or, in the more general
  case, a parallelogram similar to a given parallelogram).''

  What is meant by saying that the applied parallelogram
  (1) \emph{exceeds} or (2) \emph{falls short} is that, while its
  base coincides and is coterminous \emph{at one end} with the
  straight line, the said base (1) overlaps or (2) falls short of
  the straight line \emph{at the other end}, and the portion by
  which the applied parallelogram exceeds a parallelogram of the
  same angle and height on the given straight line (exactly) as
  base is a parallelogram similar to a given parallelogram
  (or, in particular cases, a square). In the case where the
  parallelogram is to \emph{fall short}, some such remark
  as Woepcke's (note 104) is necessary to express the condition
  of possibility of solution. For the other case see note 116.

  The solution of the problems here stated is equivalent to the
  solution of a quadratic equation. By means of
  \textsc{ii}.~5 and 6 we can solve the equations
  \begin{align*}
    ax \pm x^2 &= b^2,\\
    x^2 - ax &= b^2,
  \end{align*}
  but in \textsc{vi}.~28,~29 Euclid gives the equivalent of
  the solution of the general equations \[ax \pm px^2 = A. \]
  \textsc{vi}.~28 is:
  \textit{To a given straight line to apply a parallelogram equal
  to a given rectilineal figure and deficient by a parallelogrammic
  figure similar to a given one:
  thus the given rectilineal figure must not be greater than the parallelogram
  described on the half of the straight line and similar to the defect.}

  The Proposition~18 of Euclid under consideration is a particular case
  of this problem and as the fragment of the text and Woepcke's note
  (note~104) are contained in it, doubt may well be entertained as to
  whether Euclid gave any construction in his book \textit{On Divisions}.
  The problem can be solved without the aid of Book \textsc{vi} of the
  \textit{Elements} and by means of \textsc{ii}.~5 and
  \textsc{ii}.~14 only, as indicated in the text above.

  The appropriation of the terms parabola (\emph{application}),
  hyperbola (\emph{exceeding}) and ellipse (\emph{falling-short})
  to conic sections was first introduced by Apollonius as expressing
  in each case the fundamental properties of curves as stated by him.
  This fundamental property is the geometrical equivalent of the
  Cartesian equation referred to any diameter of the conic and the
  tangent at its extremity as (in general, oblique) axes.
  More particulars in this connection are given by Heath.

  The terms ``parabolic,'' ``hyperbolic'' and ``elliptic,''
  introduced by Klein for the three main divisions of Geometry,
  are appropriate to systems in which a straight angle equals,
  exceeds and falls short of the angle sum of any triangle.
  \textit{Cf.}\ \textsc{W.~B.~Frankland},
  \textit{The First Book of Euclid's Elements with a Commentary
  based principally upon that of Proclus Diadochus\dots}Cambridge, 1905, p.~122.
  }                                                                       % end of footnote 103

% -----File: 063.png---Folio 51------------------------------------------------------

``After having done what was required, if some one ask,
How is it possible to apply to the line $\mathit{AB}$ a rectangle such
that the rectangle $\mathit{AE}\centerdot\mathit{EB}$
\CenteredGraphic{85mm}{063}                                               % Graphic in Proposition 18
is equal to the
rectangle $\mathit{AB}\centerdot\mathit{AC}$
and deficient by a square---we say that it is impossible, because
$\mathit{AB}$ is greater than $\mathit{BE}$ and $\mathit{AC}$ greater
than $\mathit{AE}$, and consequently
the rectangle $\mathit{BA}\centerdot\mathit{AC}$ greater than the rectangle
$\mathit{AE}\centerdot\mathit{EB}$.
Then when one applies to the line $\mathit{AB}$ a parallelogram
equal to the rectangle $\mathit{AB}\centerdot\mathit{AC}$ the rectangle
$\mathit{AZ}\centerdot\mathit{ZB}$ is\dots\dots\footnotemark[104].''

\SmallLeftNote{
In this problem it is required to find in the given line $\mathit{AB}$ a
point $Z$ such that
\[
\mathit{AB}\centerdot\mathit{ZB} - \mathit{ZB}^2
 [= \mathit{AZ}\centerdot\mathit{ZB} \text{ by \textsc{ii}.~3; cf.~x.~16 lemma}] =
 \mathit{AB}\centerdot\mathit{AC}\footnotemark[105].
\]

Find, by \textsc{ii}.~14, the side, $b$, of a square equal in area
to the rectangle $\mathit{AB}\centerdot\mathit{AC}$, then the problem
is exactly equivalent to that of which a simple solution was given
by Simson\footnotemark[106]:}


\Footnotetext{104}{                                                       % Footnote 104
  Woepcke here remarks: ``Evidently if $a$ denote the length of the
  line to which the rectangle is to be applied, Problem 18 is only
  possible when
  \[\mathit{AB}\centerdot\mathit{AC} < \left( \dfrac{a}{2} \right)^2.\]
  Then if $a$ be taken as $\mathit{AB}$ one of the two sides of the
  given rectangle, relatively to the other side, $AC < \dfrac{AB}{4}$.
  It is probably the demonstration of this which was
  given in the missing portion of the text.''}                            % end footnote 104

% This footnote was labelled 106 in the 1915 Cambridge edition. Typo corrected.
\Footnotetext{\DPtypo{106}{105}}{                                         % Footnote 105
  If $\mathit{AB} = a$, $\mathit{ZB} = x$,
  $\mathit{AB}\centerdot\mathit{AC} = b^2$, the problem is to find
  a geometric solution of the equation $ax-x^2=b^2$.
  Ofterdinger\footnotemark[38] (p.~15) seems to have quite missed the
  meaning of this problem. He thought, apparently, that it was
  equivalent to \textsc{x}.~16, lemma, of the \textit{Elements}.}         % end footnote 105

\Footnotetext{106}{                                                       % Footnote 106
  \textsc{R.~Simson}, \textit{Elements of Euclid},
  ninth~ed., Edinburgh, 1793, pp.~335--6.}                                % end footnote 106
% -----File: 064.png---Folio 52------------------------------------------------------

\small
To apply a rectangle which shall be equal to a given square, to a given
straight line, deficient by a square: but the given square must not be greater
than that upon the half of the given line.

\CenteredGraphic{95mm}{064}                                               % Graphic in Proposition 18

Bisect $\mathit{AB}$ in $D$, and if the square on $\mathit{AD}$ be equal to the square on $b$,
the thing required is done. But if it be not equal to it, $\mathit{AD}$ must be greater
than $b$ according to the determination. Then draw $\mathit{DO}$ perpendicular to $\mathit{AB}$
and equal to $b$; produce $\mathit{OD}$ to $N$ so that $\mathit{ON}=\mathit{DB}$
(or $\tfrac{1}{2}a$); and with $O$ as
centre and radius $\mathit{ON}$ describe a circle cutting $\mathit{DB}$ in $Z$.

Then $\mathit{ZB}$ (or $x$) is found, and therefore the required rectangle $\mathit{AH}$.

For the rectangle $\mathit{AZ}\centerdot\mathit{ZB}$ together with the square
on $\mathit{DZ}$ is equal to the
square on $\mathit{DB}$, \hfill[\textsc{ii}.~5]\quad

\hspace{10em}i.~e.\ to the square on $\mathit{OZ}$,

\hspace{10em}i.~e.\ to the squares on $\mathit{OD, DZ}$. \hfill [\textsc{i}.~47]\quad

\noindent
Whence the rectangle $\mathit{AZ}\centerdot\mathit{ZB}$ is equal to the square on $\mathit{OD}$.

Wherefore the rectangle $\mathit{AH}$ equals the given square upon $b$ (i.~e.\ the
rectangle $\mathit{AB}\centerdot\mathit{AC}$) and has been applied to the
given straight line $\mathit{AB}$,
deficient by the square $\mathit{HB}$\footnotemark[107].

\Footnotetext{107}{                                                       % Footnote 107
  It is not in the manner of Euclid to take account of the two solutions
  found by considering $(F)$, as well as $Z$, determined by the circle
  with centre $O$.

  Although Leonardo's construction for Problem 19 is identical with that
  of Euclid who makes use of Problem 18, Leonardo does not seem to have
  anywhere formulated Problem 18. He may have considered it
  sufficiently obvious from \textsc{vi}.~28, or from \textsc{ii}.~5 and
  \textsc{ii}.~6, of which he gives the enunciations in the early
  pages (15--16) of his \textit{Practica Geometriae}; he also
  considers (p.~60) the roots of a resulting quadratic equation,
  $ax-x^2 = b$ (\textit{cf.}\ \textsc{Cantor}, \textit{Vorlesungen\dots},
  \textsc{ii}$_2$, 39), but does not give \textsc{ii}.~14.
  \textit{Cf.\ Bibliotheca Mathematica}, (3), 1907--8,
  \textsc{viii},~190; and also \textsc{ix},~245.}                         % end footnote 107


\normalsize

\Proposition{19}                                                          % Proposition 19
\Paragraph{}{40} \textit{``To divide a given triangle into two equal      % Paragraph 40
parts by a line which passes through a point situated in the
interior of the triangle.''}
[Leonardo~3, p.~115, ll.~7--10.]

% -----File: 065.png---Folio 53------------------------------------------------------

``Let the given triangle be $\mathit{ABC}$, and the given point in the interior
of this triangle, $D$.

\WrappedGraphic{50mm}{065}{15}{r}                                         % Graphic in Proposition 19

It is required to draw through
$D$ a straight line which divides
the triangle $\mathit{ABC}$ into two equal
parts.

Draw from the point $D$ a line
parallel to the line $\mathit{BC}$, as $\mathit{DE}$, and\DPnote{yes, starting a new paragraph}

Apply to $\mathit{DE}$ a rectangle equal
to half of the rectangle $\mathit{AB}\centerdot\mathit{BC}$,
such as
\[
\mathit{TB}\centerdot\mathit{DE}
\left[\mathit{TB}=\dfrac{\mathit{AB}\centerdot\mathit{BC}}{2\mathit{DE}}\right].
\]

Apply to the line $\mathit{TB}$ a
parallelogram equal to the rectangle $\mathit{BT}\centerdot\mathit{BE}$
and deficient by a square\footnotemark[1071].\hfill [Prop.~18] \qquad
% We cannot use alphanumeric footnote marker 107a used in original book.

Let the rectangle applied be
\[
\mathit{BH}\centerdot\mathit{HT}[(\mathit{TB-HT})\centerdot\mathit{HT}=
  \mathit{TB}\centerdot\mathit{BE}].
\]

Draw the line $\mathit{HD}$ and produce it to $Z$.

Then this is the line required and the triangle $\mathit{ABC}$ is
divided into two equal parts $\mathit{HBZ}$ and $\mathit{HZCA}$.

\textit{Demonstration}.
The rectangle $\mathit{TB}\centerdot\mathit{BE}$ is equal to the
rectangle $\mathit{TH}\centerdot\mathit{HB}$, whence it follows that
\begin{DPalign*}
  BT: TH &= HB: BE;  \\
  \lintertext{then \textit{dividendo}\footnotemark[108]} TB: BH &= BH: HE. \\
  \lintertext{But} BH: HE &= BZ: ED;   \tag*{[\textsc{vi}.~2]} \\
  \lintertext{therefore}  TB: BH &= BZ: ED.
\end{DPalign*}
Consequently the rectangle $\mathit{TB}\centerdot\mathit{ED}$ is equal
to the rectangle $\mathit{BH}\centerdot\mathit{BZ}$.
But the rectangle $\mathit{TB}\centerdot\mathit{ED}$ is equal to half the
rectangle $\mathit{AB}\centerdot\mathit{BC}$; and
\[\mathit{BH}\centerdot\mathit{BZ:AB}\centerdot\mathit{BC}
= \triangle\mathit{HBZ}: \triangle\mathit{ABC}\footnotemark[88],
\]
% -----File: 066.png---Folio 54------------------------------------------------------
since the angle $B$ is common. The triangle $\mathit{HBZ}$ is, then,
half the triangle $\mathit{ABC}$.

\Footnotetext{1071}{                                                      % Footnote 1071 (was 107a)
  The corresponding sentence in Leonardo is (p.~115, ll.~15--17):
  ``Deinde linee $\mathit{gz}$ applicabis paralilogramum deficiens
  figura tetragona, quod sit equale
  superficies $\mathit{ge}$ in $\mathit{gz}$.''}                          % end footnote 1071
\Footnotetext{108}{                                                       % Footnote 108
  ``Elements, Book~\textsc{v}, definition~16'' (Woepcke).
  This is definition~15 in \textsc{Heath},
  \textit{The Thirteen Books of Euclid's Elements},
  \textsc{ii},~135.}                                                      % end footnote 108

Therefore the triangle $\mathit{ABC}$ is divided into two equal
parts $\mathit{BHZ}$ and $\mathit{AHZC}$.

If, in applying to $\mathit{TB}$ a parallelogram equal to the rectangle
$\mathit{TB}\centerdot\mathit{BE}$ and of which the complement
is a square, we obtain the rectangle
$\mathit{AB}\centerdot\mathit{AT}$\Footnote{109}{                         % Footnote 109
  ``In other words when $H$ coincides with $A$.
  This can only be the case
  when $D$ is situated on the line which joins $A$ to the middle of
  the base $\mathit{BC}$''
  (Woepcke). If $D$ were at the centre of gravity of the triangle,
  three lines could be
  drawn through $D$ dividing the triangle into two equal parts.
  As introductory to his Prop.~3, Leonardo proved that the medians of a
  triangle meet in a point, and trisect one another---results known
  to Archimedes\footnotemark[601],\DPnote{* was 60a}
  but no complete, strictly geometric proof has come down to us from
  the Greeks. Leonardo then proves that if a point be taken on any
  one of the medians, or on one of the medians produced, the line
  through this point and the corresponding angular point of the
  triangle will divide the triangle into two equal parts.
  He next shows that lines through the vertices of a triangle and
  any point within not on one of the medians, will divide the given
  triangle into triangles whose areas are each either greater than or
  less than the area of half of the original triangle.
  This leads Leonardo to the consideration of the problem, to draw
  through a point, within a triangle and not on one of the medians,
  a line which \emph{will} bisect the area of the triangle.
  (Euclid, Prop.~19.)

  The last paragraph of Euclid's proof, as it has come down to us
  through Arabian sources, does not ring true, and it was not in
  the Euclidean manner to consider special cases.

  After Leonardo's proof of Proposition~19,
  a numerical example is given.}%                                          % end footnote 109
, we may demonstrate in
an analogous manner, by drawing the line $\mathit{AD}$ and prolonging
it to $K$, that the triangle $\mathit{ABK}$ is one-half of the triangle
$\mathit{ABC}$. And this is what was required to be demonstrated.''

\Proposition{20}                                                          % Proposition 20
\Paragraph{}{41} \textit{``To cut off a certain fraction from a given     % Paragraph 41
triangle by a line drawn from a given point situated in the interior
of the triangle.''} [Leonardo~10, p.~121, ll.~1--2.]

``Let $\mathit{ABC}$ be the given triangle and $D$ the given point in the interior
of the triangle.
It is required to pass through
the point $D$ a straight line
which cuts off a certain fraction
of the triangle $\mathit{ABC}$.

\WrappedGraphic{50mm}{066}{}{r}                                           % Graphic in Proposition 20

``Let the certain fraction
be one-third. Draw from the
point $D$ a line parallel to the
line $\mathit{BC}$, as $\mathit{DE}$, and apply
to $\mathit{DE}$ a rectangle equal to
one-third of the rectangle
$\mathit{AB}\centerdot\mathit{BC}$. Let this be
\[
  \mathit{BZ}\centerdot\mathit{ED}
  \left[\mathit{BZ}=\frac{\mathit{AB}\centerdot\mathit{BC}}{3\centerdot\mathit{ED}}\right].
\]
% -----File: 067.png---Folio 55------------------------------------------------------
Then apply to $\mathit{ZB}$ a rectangle equal to the rectangle
$\mathit{ZB}\centerdot\mathit{BE}$
and deficient by a square. [Prop.~18.] Let the rectangle
applied be the rectangle
\[
\mathit{BH}\centerdot\mathit{HZ} \left[\left(ZB-HZ\right) HZ=ZB\centerdot BE\right].
\]
Draw the line $\mathit{HD}$ and produce it to $T$.

``On proceeding as above we may demonstrate that the
triangle $\mathit{HTB}$ is one-third of the triangle $\mathit{ABC}$; and by
means of an analogous construction to this we may divide
the triangle in any ratio. But this is what it is required
to do\Footnote{110}{                                                      % Footnote 110
  Leonardo gives the details of the proof for the case of one-third
  and does not refer to any other fraction.
  If, however, the ``certain fraction'' were the ratio
  of the lengths of two given lines, $m:n$, we could readily
  construct a rectangle equal to
  $\dfrac{m}{n}\centerdot AB\centerdot BC$, and then find the
  rectangle $\mathit{BZ}\centerdot\mathit{ED}$ equal to it.
  The rest of the construction is the same as given above.

  According to the conditions set forth in Proposition~18,
  there will be two, one, or no solutions of Propositions~19 and 20.
  Leonardo considers only the Euclidean cases.
  \textit{Cf.}\ notes 104 and 107.

  The case where there is no solution may be readily indicated.
  Suppose, in the above figure, that $\mathit{BE}=\mathit{EH}$, then of all triangles
  formed by lines drawn through $D$ to meet $\mathit{AB}$ and $\mathit{BC}$,
  the triangle $\mathit{HBT}$ has the minimum area. (Easily shown synthetically
  as in \textsc{D.~Cresswell},
  \textit{An Elementary Treatise on the Geometrical and
  Algebraical Investigations of Maxima and Minima}.
  Second edition, Cambridge, 1817, pp.~15--17.)
  Similar minimum triangles may be found in connection with the pairs
  of sides $\mathit{AB},\mathit{AC}$ and $\mathit{AC},\mathit{CB}$.
  Suppose that neither of these
  triangles is less than the triangle $\mathit{HBT}$. Then if
  \[\triangle\mathit{HBT}: \triangle\mathit{ABC} > m:n,\]
  the solution of the problem is impossible.}                             % end footnote 110
.''

\Proposition{21}                                                          % Proposition 21
\Paragraph{}{42} \textit{``Given the four lines $A$, $B$, $C$, $D$ and that the   % Paragraph 42
product of $A$ and $D$ is greater than the product of $B$ and $C$;
I say that the ratio of $A$ to $B$ will be greater than the ratio
of $C$ to $D$''}\Footnote{111}{                                           % Footnote 111
  This and the next four auxiliary propositions for which I supply        % a really long footnote
  possible proofs, seem to be neither formally stated nor proved by
  Leonardo.
  At least some of the results are nevertheless assumed in his
  discussion of Euclid's later propositions, as we shall presently see.
  Although these auxiliary propositions are not given in the
  \textit{Elements}, they are assumed as known by Archimedes, Ptolemy
  and Apollonius.

  For example, in Archimedes' ``On Sphere and Cylinder,''
  \textsc{ii}.~9 (Heiberg, ed.~\textsc{i}, 1910, p.~227;
  Heath, ed.\ 1897, p.~90), Woepcke~21 is used.
  See also Eutocius' Commentary (Archimedis \textit{Opera omnia}
  ed.~Heiberg, \textsc{iii}, 1881, p.~257, etc.), and
  \textsc{Heiberg}, \textit{Quaestiones Archimedeae}, Hauniae, 1879,
  p.~45~f.
  For a possible application by Archimedes (in his
  \textit{Measurement of a circle}) of what is practically equivalent
  to Woepcke~24, see Heath's \textit{Archimedes\dots}, 1897, p.~xc.

  The equivalent of Woepcke~24 is assumed in the proof of a proposition
  given by Ptolemy (87--165~A.~D.) in his \textit{Syntaxis},
  vol.~\textsc{i}, Heiberg edition, Leipzig, 1898, pp.~43--44.
  This in turn is tacitly assumed by Aristarchus of Samos (\textit{circa}
  310--230~B.~C.) in his work
  \textit{On the Sizes and Distances of the Sun and Moon}
  (see Heath's edition \textit{Aristarchus of Samos the Ancient Copernicus},
  Oxford, 1913, pp.~367, 369, 377, 381, 389, 391).

  As to the use of the auxiliary propositions in the two works
  \textit{Proportional Section} and \textit{On Cutting off a Space},
  of Apollonius, we must refer to Pappus' account
  (Pappi Alexandrini \textit{Collectionis\dots} ed.~Hultsch,
  vol.~\textsc{ii}, 1877, pp.~684 ff.).
  Woepcke~21, 22 occur on pp.~696--697; Woepcke~24 enters on pp.~684--687;
  Woepcke~23, 25 are given on pp.~687, 689. Perhaps this last statement
  should be modified; for whereas Euclid's propositions affirm that if
  \[  a:b \gtrless c:d, \quad a-b:b \gtrless c-d:d, \]
  Pappus shows that if \[ a:b \gtrless c:d, \quad a:a-b \lessgtr c:c-d; \]
  but these propositions are immediately followed by others which state
  that if \[ a:b \gtrless c:d, \text{ then } b:a \lessgtr d:c. \]

  Below is given a list of the various restorations of the above-named
  works of Apollonius, based on the account of Pappus.
  By reference to these restorations the way in which the auxiliary
  propositions are used or avoided may be observed.
  We have already (Art.~21) noticed a connection of Apollonius' work
  \textit{On Cutting off a Space} with our subject under discussion.
  Some of these titles will therefore supplement the list given in the Appendix.

  \textit{Wilebrordi Snellii R.~F.}
  \ItalicGreekSnippet{per`i l'ogou `apotomês ka`i per`i chwr'iou `apotomhs}
  \textit{(Apollonii) resuscitata geometria.} Lugodini,
  \textit{ex officina Platiniana} Raphelengii, \textsc{md.cvii} pp.~23.

More or less extensive abridgment of Snellius's work is given in:

\renewcommand{\labelenumi}{(\alph{enumi})}
\begin{enumerate}
\item \textit{Universae geometriae mixtaeque mathematicae synopsis et bini
refractionum demonstratarum tractatus. Studio et opera F.~M.\ Mersenni.}
Parisiis, \textsc{m.dc.xliv}, p.~382.
\item \textit{Cursus mathematicus}, P.~Herigone. Paris, 1634, tome~\textsc{i}, pp.~899--904;
also Paris, 1644.
\end{enumerate}

\textit{Apollonii Pergaei de sectione rationis libri duo ex Arabico $MS^{to}$ Latine versi
accedunt ejusdem de sectione spatii libri duo restituti\dots opera \& studio Edmundi
Halley}\dots Oxonii,\dots \textsc{mdccvi}, pp.~8 + liii + 168.

\begin{enumerate}
\item \textit{Die Bücher des Apollonius von Perga De sectione rationis nach dem
Lateinischen des Edm.\ Halley frey bearbeitet, und mit einem Anhange
versehen von W.~A.\ Diesterweg}, Berlin, 1824, pp.~xvi + 218 + 9~pl.
\item \textit{Des Apollonius von Perga zwei Bücher vom Verhältnissschnitt}
(\textit{de sectione rationis}) \textit{aus dem Lateinischen des Halley übersetzt und mit
Anmerkungen begleitet und einem Anhang versehen von August Richter}
\DPnote{[**parentheses ( ) should not be italicized]}
\ldots Elbing, 1836, pp.~xxii + 143 + 4~pl.
\end{enumerate}

\textit{Die Bücher des Apollonius von Perga de sectione
spatii wiederhergestellt von
Dr~W.~A.~Diesterweg\dots} Elberfeld, 1827\dots pp.~vi + 154 + 5~pl.

\textit{Des Apollonius von Perga zwei Bücher vom Raumschnitt. Ein Versuch in der
alten Geometrie von A.~Richter.} Halberstadt, 1828, pp.~xvi + 105 + 9~pl.

\textit{Die Bücher des Apollonius von Perga de sectione spatii, analytisch bearbeitet
und mit einem Anhange von mehreren Aufgaben ähnlicher Art versehen
von M.~G.~Grabow\dots} Frankfurt a.~M., 1834, pp.~80 + 3~pl.

\textit{Geometrische Analysis enthaltend des Apollonius von Perga sectio rationis,
spatii und determinata, nebst einem Anhange zu der letzten, neu bearbeitet vom
Prof.~Dr~Georg Paucker}, Leipzig, 1837, pp.~xii + 167 + 9~pl.

M.~Chasles discovered that by means of the theory of involution a single
method of solution could be applied to the main problem of the three books of
Apollonius above mentioned. This solution was first published in \textit{The Mathematician},
vol.~\textsc{iii}, Nov.~1848, pp.~201--202. This is reproduced by A.~Wiegand in
his \textit{Die schwierigeren geometrischen Aufgaben aus des Herrn Prof.~C.~A.~Jacobi
Anhängen zu Van Swinden's Elementen der Geometrie. Mit Ergänzungen
englischer Mathematiker\dots} Halle, 1849, pp.~148--149, and it appears at greater
length in Chasles' \textit{Traité de Géométrie supérieure},
Paris, 1852, pp.~216--218; $2^\text{e}$ éd.
1880, pp.~202--204. It was no doubt Chasles who inspired \textit{Die Elemente der
projectivischen Geometrie in synthetischer Behandlung. Vorlesungen von H.~Hankel},
(Leipzig, 1875), ``Vierter Abschnitt, Aufgaben des Apollonius,'' pp.~128--145;
``sectio rationis,'' pp.~128--138; ``sectio spatii,'' pp.~138--140.

The \textit{``Three Sections,'' the ``Tangencies'' and a ``Loci
  Problem'' of Apollonius\ldots
by M.~Gardiner}, Melbourne, 1860. Reprinted from the \textit{Transactions of the Royal
Society of Victoria}, 1860--1861, \textsc{v},~19--91 + 10~pl.

\textit{Die sectio rationis, sectio spatii und sectio determinata des Apollonius nebst
einigen verwandten geometrischen Aufgaben von Fr.\ von Lühmann}. Progr.\
Königsberg in d.~N.~1882, pp.~16 + 1~pl.

  ``Ueber die fünf Aufgaben des Apollonius,''
  \textit{von L.~F.~Ofterdinger. Jahreshefte des Vereines
  für Math.\ u.\ Naturwiss.\ in Ulm a.~D.}\ 1888, \textsc{i},~21--38;
  ``Verhältnissschnitt,'' pp.~23--25;
  ``Flächenschnitt,'' pp.~26--27.}%                                       % end footnote 111 (whew!)
.   % this period is the end of the sentence pointing to footnote 111 (140 lines back!)

% -----File: 068.png---Folio 56------------------------------------------------------

\normalsize
\small
Given $A\centerdot D > B\centerdot C$. To prove $A:B > C:D$.

Let the lines $A$, $D$ be adjacent sides of a rectangle; and let there be
another rectangle with side $B$ lying along $A$ and side $C$ along $D$. Then
either $A$ is greater than $B$, or $D$ greater than $C$, for otherwise the rectangle
$A\centerdot D$ would not be greater than the rectangle $B\centerdot C$.

% -----File: 069.png---Folio 57------------------------------------------------------
Let then $A > B$. To $D$ apply the rectangle $B\centerdot C$ and we get a rectangle
\begin{DPalign*}
 A'\centerdot D &= B\centerdot C;  \tag*{[\textsc{i}.~44--45] \quad} \\
\lintertext{then}  A':B &= C:D.    \tag*{\text{[\textsc{vii}.~19] \quad}} \\
\lintertext{\indent But since }   A &> A',  \\
 A:B &> A':B;  \tag*{\text{[\textsc{v}.~8] \quad}} \\
 \therefore A:B &> C:D.   \tag*{\text{[\textsc{v}.~13] \quad}}\\
 \tag*{\textsc{q.~e.~d.}\quad\quad}
\end{DPalign*}

Pappus remarks: Conversely if $A:B > C:D$, $A\centerdot D > B\centerdot C$. The proof
follows at once.
\begin{DPalign*}
 \lintertext{\indent For, find $A'$ such that}  A': B &= C:D; \\
 \lintertext{then}   A:B &> A':B,
\end{DPalign*}
and $A>A'$. But $A'\centerdot D = B\centerdot C.\quad
\therefore A\centerdot D > B\centerdot C.$  \hfill \textsc{q.~e.~d.} \qquad\qquad
\normalsize

\Proposition{22}                                                          % Proposition 22
\Paragraph{}{43} \textit{``And when the product of $A$ and $D$ is less than   % Paragraph 43
the product of $B$ and $C$, then the ratio of $A$ to $B$ is less than the
ratio of $C$ to $D$.''}


% -----File: 070.png---Folio 58------------------------------------------------------
\SmallLeftNote{
From the above proof we evidently have
\[C:D > A:B,\]
that is,
\[A:B < C:D.\]
Conversely, as above, if $A:B < C:D$, $A\centerdot B < C\centerdot D$.

It is really this converse, and not the proposition, which Euclid uses in
Proposition~26. Proclus remarks (page~407) that the converses of Euclid's
Elements, \textsc{i}.~35,~36, about parallelograms, are unnecessary ``because it is easy
to see that the method would be the same, and therefore the reader may
properly be left to prove them for himself.'' No doubt similar comment is
justifiable here.}

\Proposition{23}                                                          % Proposition 23
\Paragraph{}{44} \textit{``Given any two straight lines and on these      % Paragraph 44
lines the points $A$,~$B$, and $D$,~$E$; and let the ratio of
$\mathit{AB}$ to $\mathit{BC}$ be greater than the ratio of
$\mathit{DE} : \mathit{EZ}$;}
\textit{I say that \emph{dividendo} the
ratio of $\mathit{AC}$ to $\mathit{CB}$ will be greater than the ratio of
$\mathit{DZ}$ to $\mathit{ZE}$.''}

\CenteredGraphic{90mm}{070}                                               % Graphic in Proposition 23

\small
\begin{DPgather*}
 \lintertext{\indent Given}      AB:BC > DE:EZ.\\
 \lintertext{\indent To prove }  AC:CB > DZ:ZE.
\end{DPgather*}
\indent To $\mathit{AB}, \mathit{BC}, \mathit{DE}$ find a fourth proportional $\mathit{EW}$. \hfill    [\textsc{vi.}~12]

\begin{DPgather*}
 \lintertext{\indent Then} AB:BC = DE:EW. \tag*{\ldots\ldots\ldots(1)}\\
 \lintertext{\indent But } AB:BC > DE:EZ; \\
 \therefore DE:EW > DE:EZ.\tag*{[\textsc{v.}~13]} \\
 \therefore EW < EZ.  \tag*{[\textsc{v.}~8]} \\
 \lintertext{\indent From (1)} AC:CB = DW:WE; \tag*{\ldots\ldots\ldots(2)~[\textsc{v.}~17]}\\
 \lintertext{\indent since} DW>DZ,\quad DW:WE > DZ:WE. \tag*{[v.~8]} \\
  \therefore AC:CB > DZ:WE. \tag*{[\textsc{v.}~13]} \\
 \lintertext{\indent But} WE < ZE; \quad\therefore DZ:WE > DZ:ZE. \tag*{[\textsc{v.}~8]} \\
  \therefore AC:CB > DZ:ZE. \tag*{From~(2)~and~[\textsc{v.}~17]}\\
  \tag*{\textsc{q.~e.~d.}\quad}
\end{DPgather*}

\normalsize

% -----File: 071.png---Folio 59------------------------------------------------------
\Proposition{24}                                                          % Proposition 24
\Paragraph{}{45} \textit{``And in an exactly analogous manner I say that   % Paragraph 45
when the ratio of $\mathit{AC}$ to $\mathit{CB}$ is greater than the ratio of
$\mathit{DZ}$ to $\mathit{ZE}$, we shall have} componendo\Footnote{112}{                           % Footnote 112
  ``Elements, Book \textsc{v}, definition 15'' (Woepcke).
  This is definition 14 in \textsc{Heath},
  \textit{The Thirteen Books of Euclid's Elements}, \textsc{ii}, 135.}    % end footnote 112
\textit{the ratio of $\mathit{AB}$ to $\mathit{BC}$ is
greater than the ratio of $\mathit{DE}$ to $\mathit{EZ}$.''}

\small
\begin{DPgather*}
 \lintertext{\indent Given}      AC:CB > DZ:ZE. \\
 \lintertext{\indent To prove}     AB:BC > DE:EZ.
\end{DPgather*}
 \indent Determine $W$, as before, such that
\begin{DPgather*}
  AB:BC = DE:EW.  \\
 \lintertext{\indent Then}  AC:CB = DW:WE.    \tag*{[\textsc{v}. 17]} \\
 \therefore \; DW:WE > DZ:ZE. \tag*{\ldots\ldots (1) \qquad  [\textsc{v}. 13]} \\
 \lintertext{\indent Now either}    EW>EZ \text{ or } EW<EZ.
\end{DPgather*}
\indent If $EW>EZ$,\quad $DW<DZ$, and
\begin{DPalign*}
 DW:EW&<DZ:EW.   \tag*{[\textsc{v}. 8]} \\
 \lintertext{\indent So much the more is} & \\
  DW:EW&<DZ:EZ   \tag*{ [\textsc{v}. 8]} \\
\lintertext{which contradicts (1).}
\end{DPalign*}
\begin{DPalign*}
 \therefore  EW &< EZ. \\
 \lintertext{\indent But}        AB:BC &= DE:EW, \\
 \lintertext{and}        DE:EW &> DE:EZ;  \tag*{[\textsc{v}. 8]} \\
 \therefore AB:BC &> DE:EZ.    \tag*{ [\textsc{v}. 13]} \\
 & \tag*{\textsc{q.~e.~d.}\quad}
\end{DPalign*}

\iffalse
\begin{math}
\begin{array}{llr}
\text{\indent Given}   &   \mathit{AC:CB}>\mathit{DZ:ZE}.& \\
\text{\indent To prove}  &   AB:BC>DE:EZ. & \\
\multicolumn{3}{l}{\text{\indent Determine $W$, as before, such that}} \\
          & AB:BC=DE:EW. & \\
\text{\indent Then}      & AC:CB=DW:WE.    &    [\textsc{v}. 17] \\
& \therefore DW:WE>DZ:ZE. \ldots\ldots (1) &  [\textsc{v}. 13] \\
\text{\indent Now either} &   EW>EZ\text{ or }EW<EZ. & \\
\multicolumn{3}{l}{\text{\indent If }EW>EZ,\quad DW<DZ, \text{ and}} \\
& DW:EW<DZ:EW. &      [\textsc{v}. 8]\\
\multicolumn{3}{l}{\text{\indent So much the more is}} \\
& DW:EW<DZ:EZ &      [\textsc{v}. 8] \\
\multicolumn{3}{l}{\text{which contradicts (1).}} \\
& \therefore  EW<EZ. \\
\text{\indent But}  &      AB:BC=DE:EW, \\
\text{and}  &      DE:EW>DE:EZ; &      [\textsc{v}. 8] \\
& \therefore AB:BC>DE:EZ. &      [\textsc{v}. 13] \\
& & \textsc{q.~e.~d.}\quad
\end{array}
\end{math}
\fi
\normalsize

\Proposition{25}                                                          % Proposition 25
\Paragraph{}{46} \textit{``Suppose again that the ratio of                % Paragraph 46
$\mathit{AB}$ to $\mathit{BC}$ were less than the}
\CenteredGraphic{85mm}{071}                                               % Graphic in Proposition 25
\textit{ratio of $\mathit{DE}$ to $\mathit{EZ}$; \emph{dividendo}
the ratio of $\mathit{AC}$ to $\mathit{CB}$ will be less than the
ratio of $\mathit{DE}$ to $\mathit{ZE}$}\Footnote{113}{                                         % Footnote 113
  The auxiliary propositions are introduced, apparently, to assist
  in rendering, with faultless logic, the remarkable proof of
  Proposition 26. In this proof it will be observed that we are
  referred back to Proposition 21, to the converse of Proposition 22
  and to Proposition 25 only, although 23 is really the same as 25.
  But no step in the reasoning has led to Proposition 24. If this
  is unnecessary, why has it been introduced?

  To answer this question, let us inspect the auxiliary propositions
  more closely. In a sense Propositions~21 and 22 go together:
  If $ad \gtrless bc$, then $a:b \gtrless c:d$.
  So also for Propositions~23 and 25:
  If $a:b \gtrless c:d$, then $a-b:b \gtrless c-d:d$.
  Proposition~24 is really the converse of 23:
  If $a:b > c:d$, then $a+b:b > c+d:d$.
  Had Euclid given another proposition:
  If $a:b < c:d$ then $a+b:b < c+d:d$,
  we should have had two groups of propositions 21, 22, and 23, 25
  with their converses. Now the converses of 21 and 22 are exceedingly
  evident in both statement and proof. But this can hardly be said
  of the proof of 24, the converse of 23. The converse of 23 having
  been given the formulation of the statement and proof of the
  converse of 25 is obvious and unnecessary to state, according
  to Euclid's ideals (cf.\ Art.~43).
  It might therefore seem that Proposition~24 is merely given to
  complete what is not altogether obvious, in connection with the
  statement of the four propositions 21 and 22, 23 and 25, and their
  converses. In Pappus' discussion some support is given to this view,
  since Propositions 21 and 22 and converses are treated as a single
  proposition; Propositions 23, 25 as another proposition, while the
  converses of 23 and 25 are dealt with separately.

  The more probable explanation is, however, that Propositions 23 and
  24 were given by Euclid because they were necessary for the
  discussion of other cases of Proposition~26 (assuming that the first
  case of Leonardo was that given by Euclid), for it was not his
  manner to consider different cases. Indeed if we take $\mathit{be}$ less than
  $\mathit{ge}$ in the first part of Leonardo's discussion exactly Propositions
  23 and 24 are necessary.}.''                                            % end of footnote 113

% -----File: 072.png---Folio 60------------------------------------------------------
\SmallLeftNote{Just as the proof of Proposition~22 was contained in that for Proposition~21,
so here, the proof required is contained in the proof of Proposition~23.
Similarly the converse of Proposition~25 flows out of 24.}

\Proposition{26}                                                          % Proposition 26
\Paragraph{}{47} \textit{``To divide a given triangle into two equal      % Paragraph 47
parts by a line drawn from a given point situated outside the triangle.''}
[Leonardo 4, p.~116, ll.~35--36.]

Let the triangle be $\mathit{abg}$ and $d$ the point outside.

Join $\mathit{ad}$ and let $\mathit{ad}$ meet $\mathit{bg}$ in $e$.
If $\mathit{be}=\mathit{eg}$, what was required is done.
For the triangles $\mathit{abe, aeg}$ being on equal
bases and of the same altitude are equal in area.

But if $\mathit{be}$ be not equal to $\mathit{eg}$, let it be greater,
and draw through $d$, parallel to $\mathit{bg}$,
a line meeting $\mathit{ab}$ produced in $z$.
\begin{flalign*}
  &\text{\indent Since } &  \mathit{be} &> \tfrac{1}{2}\mathit{bg},  && \\
  &&  \text{area } \mathit{ab}\centerdot\mathit{be} &>
    \tfrac{1}{2}\:\text{area } \mathit{ab}\centerdot\mathit{bg};
    & \text{[\textit{Cf.}\ \textsc{vii.}~17]}  &\\
  &\text{much more then is}&&&&\\
  && \text{area } \mathit{ab}\centerdot\mathit{zd}
    &> \tfrac{1}{2}\:\text{area } \mathit{ab}\centerdot\mathit{bg},
    & \text{since } \mathit{zd} > \mathit{be}.  &
\end{flalign*}
\begin{flalign*}
  &\text{\indent Now take }&&&&\\
  &&  \text{area } \mathit{ib}\centerdot\mathit{zd}
    &= \tfrac{1}{2}\:\text{area } \mathit{ab}\centerdot\mathit{bg};
    & \text{[\textsc{i.}~44]}  & \\
  &\text{then }
    & \text{area } \mathit{ab}\centerdot\mathit{be} &>
    \text{ area } \mathit{ib}\centerdot\mathit{zd},  && \\
  &\text{and }
    & \mathit{zd}:\mathit{be} &< \mathit{ba}:\mathit{bi}\footnotemark[114].
    & \text{[Prop.~21 or 22]} &
\end{flalign*}
\begin{flalign*}
  &\text{\indent But} & \mathit{zd}:\mathit{be}&=\mathit{za}:\mathit{ab}, & [\textsc{vi}.\;4]\quad& \\
  && \therefore \mathit{zb:ba} &< \mathit{ai:ib}; & \textsc{[v.}\ \text{13 and Prop. 25]}& \\
  &\text{or} & \text{area } \mathit{zb}\centerdot\mathit{bi} &<
           \text{area } \mathit{ba}\centerdot\mathit{ai}. & \text{[Converse of Prop. 22]}&
\end{flalign*}

\Footnotetext{114}{                                                       % Footnote 114
  Therefore $\mathit{bi}<\mathit{ba}$,
  and if $\mathit{bi}$ be measured along $\mathit{ba}$,
  $i$ will fall between $b$ and $a$.}                                     % end footnote 114

% -----File: 073.png---Folio 61------------------------------------------------------

Apply a rectangle equal to the rectangle
$\mathit{zb}\centerdot\mathit{bi}$ to the line $\mathit{bi}$,
but exceeding by a square\Footnote{115}{                                  % Footnote 115
  Here again we have an expression with the true Greek ring:
  ``adiungatur quidem recte
  $\centerdot\mathit{bi}\centerdot$
  paralilogramum superhabundans figura tetragona equale superficiei
  $\centerdot\mathit{zb}\centerdot$ in
  $\centerdot\mathit{bi}\centerdot$''};                                   % end footnote 115
that is to $\mathit{bi}$ apply a line such
that when multiplied by itself and by $\mathit{bi}$ the sum will be
equal to the product of $\mathit{zb}$ and $\mathit{bi}$;
let $\mathit{ti}$ be the side of the square\Footnote{116}{                % Footnote 116
  We we have seen that $i$ lies between $b$ and $a$.
  And since it has been shown that
  $\mathit{zb}\centerdot\mathit{bi} < \mathit{ba}\centerdot\mathit{ai}$,
  we now have
  $\mathit{ba}\centerdot\mathit{ai} > \mathit{bt}\centerdot\mathit{ti}$.
  If $\mathit{bt}>\mathit{ba}$, $\mathit{ti}$ is also greater than $\mathit{ai}$,
  and $\mathit{bt}\centerdot\mathit{ti} \nless \mathit{ba}\centerdot\mathit{ai}$.
  Therefore $\mathit{bt}<\mathit{ba}$ and $t$ falls between $b$ and $a$.
  But it also falls between $a$ and $i$ by reason of the construction
  (always possible) which is called for.

  In his book on \textit{Divisions} (\textit{of figures})
  Euclid does not
  formulate the proposition here quoted, possibly because of its
  similarity to Proposition 18 (see note 103).

\CenteredGraphic{70mm}{073}                                               % Graphic in Proposition 26

  If we let the rectangle $\mathit{zb}\centerdot\mathit{bi} = c^2$,
  $\mathit{ti} = x$, and $\mathit{bi} = a$,
  we have to solve geometrically the quadratic equation:
  \[
  ax+x^2=c^2.
  \]

  Heath points out (\textit{Elements}, vol.\ \textsc{i}, pp.~386--387)
  that the solution of a problem theoretically equivalent to the
  solution of a quadratic equation of this kind is presupposed in
  the fragment of Hippocrates' \textit{Quadrature of lunes}
  (5th century~\textsc{b.~c.})\ preserved in a quotation by Simplicius
  (fl.~500~\textsc{a.~d.}) from Eudemus' \textit{History of Geometry}
  (4th century \textsc{b.~c.}). See Simplicius'
  \textit{Comment.\ in Aristot.\ Phys.} ed.\ H.~Diels, Berlin, 1882,
  pp.~61--68; see also \textsc{F.~Rudio},
  \textit{Der Bericht des Simplicius über die Quadratur des
  Antiphon und Hippokrates}, Leipzig, 1907.

  Moreover as Proposition 18 is suggested by the \textit{Elements},
  \textsc{ii}.\ 5, so here this problem is suggested by \textsc{ii}.\ 6:
  \textit{If a straight line be bisected and a straight line be
  added to it in a straight line, the rectangle contained by the
  whole with the added straight line and the added straight line
  together with the square on the half is equal to the square on
  the straight line made up of the half and the added straight line.}



  If $\mathit{AB}$ is the straight line bisected at $C$ and $\mathit{BD}$ is the
  straight line added, then by \textsc{ii}.~6,
\[
  AD\centerdot DB + CB^2 = CD^2.
\]

  In his solution of our problem, Robert Simson proceeds, in effect,
  as follows (\textit{Elements of Euclid}, ninth ed., Edinburgh,
  1793, p.~336): Draw $\mathit{BQ}$ at right angles to $\mathit{AB}$ and equal to $c$.
  Join $\mathit{CQ}$ and describe a circle with centre $C$ and radius $\mathit{CQ}$
  cutting $\mathit{AB}$ produced in $D$.
  Then $\mathit{BD}$ or $x$ is found. For, by \textsc{ii}.~6,
  \begin{flalign*}
   && AD\centerdot DB + CB^2 &= CD^2,&& \\
   &&                        &= CQ^2,&& \\
   &&                        &= CB^2 + BQ^2, &&\\
   && \therefore AD\centerdot DB &= BQ^2,&&\qquad \\
   &\text{whence} & (a+x)x &= c^2&& \\
   &\text{or} & ax+x^2 &= c^2. &&
  \end{flalign*}
  It was not Euclid's manner to consider more than one solution
  in this case.
}.                                                                        % end footnote 116

Draw the straight line $\mathit{tkd}$. Since

\WrappedGraphic{40mm}{074}{}{r}                                           % Graphic in Proposition 26
\[
\text{area }\mathit{zb}\centerdot\mathit{bi}=
   \mathit{bi}\centerdot\mathit{ti} + \mathit{ti}^2 =
   \text{area }\mathit{bt}\centerdot\mathit{ti},
\]

% -----File: 074.png---Folio 62------------------------------------------------------

\begin{flalign*}
  &          & zb:bt &=ti:ib, & \textsc{[vii.~19]}& \\
  &\text{or} & zt:bt &=bt:bi. & \textsc{[v.~18]}& \\
  &\text{But}& zt:bt &=zd:bk, & \textsc{[vi.~4]}& \\
  &          & \therefore zd:bk &=bt:bi, && \\
  &\text{and}& \text{area } kb\centerdot bt &= \text{area } zd\centerdot bi. && \\
  &\text{But}& \text{area } zd\centerdot bi&= \tfrac{1}{2}\:\text{area } ab\centerdot bg, && \\
  & & \therefore \triangle\mathit{tbk} &= \tfrac{1}{2}\:\triangle\mathit{abg}\footnotemark[88]. & \\
\end{flalign*}


Therefore the triangle $\mathit{abg}$ is
divided by a line drawn from the
point $d$, that is, by the line $\mathit{tkd}$, into
two equal parts one of which is
the triangle $\mathit{tbk}$, and the other the quadrilateral $\mathit{tkga}$.

\hfill \textsc{q.~e.~f.}\qquad\qquad

\SmallLeftNote{Leonardo now gives a numerical example. He then continues:}
% -----File: 075.png---Folio 63------------------------------------------------------

[If the point $d$ were on one
side, $\mathit{ab}$, produced at say, $z$],
through $z$ draw $\mathit{ze}$ parallel to
$\mathit{bg}$ and meeting $\mathit{ag}$ produced
in $e$.

\WrappedGraphic{45mm}{175}{}{r}                                           % Graphic in Proposition 26

Make
\[
\text{area } \mathit{ze}\centerdot\mathit{gi} =
   \tfrac{1}{2}\:\text{area } \mathit{ag}\centerdot\mathit{gb},
\]
and apply a rectangle, equal to
the rectangle $\mathit{eg}\centerdot\mathit{gi}$, to the line
$\mathit{gi}$, but exceeded by a square;
\begin{DPgather*}
  \lintertext{then} \mathit{eg}\centerdot\mathit{gi}=\mathit{gt}\centerdot\mathit{ti}.
\end{DPgather*}

Join $\mathit{tz}$, then [this is the required line. The proof is step
for step as in the first case].

\SmallLeftNote{Leonardo then remarks: ``Que etiam demonstrentur in numeris,'' and
proceeds to a numerical example. Thereafter he continues:}

\WrappedGraphic{50mm}{275}{}{r}                                           % Graphic in Proposition 26

But let the sides $\mathit{ab},\mathit{gb}$ of the triangle be produced to $d$ and $e$
respectively; and let $i$
be the given point in the angle
$\mathit{ebd}$ from which a line is to be
drawn dividing the triangle
$\mathit{abg}$ into two equal parts. Join
$\mathit{ib}$ and produce it to meet $\mathit{ag}$ in
$z$. If $\mathit{az}=\mathit{zg}$, the triangle $\mathit{abg}$
is divided into two equal parts
by the line $\mathit{iz}$. But [if $\mathit{az}>\mathit{zg}$,]
let $\mathit{za}$ produced meet, in the
point $t$, the line drawn through
$i$ parallel to $\mathit{ab}$.

\begin{DPgather*}
 \lintertext{\indent Since} \mathit{za}>\tfrac{1}{2}\mathit{ag},
  \qquad\text{area } \mathit{ab}\centerdot\mathit{az} >
  \tfrac{1}{2}\:\text{area } \mathit{ba}\centerdot\mathit{ag}. \\
\lintertext{Make}  \text{area } \mathit{it}\centerdot\mathit{ak} =
 \tfrac{1}{2}\text{area }\mathit{ba}\centerdot\mathit{ag}; \\
\lintertext{then make}  \text{area } \mathit{al}\centerdot\mathit{kl}=
 \text{area } \mathit{ta}\centerdot\mathit{ak}.
\end{DPgather*}

Join $\mathit{il}$. Then as above the triangle $\mathit{abg}$ is divided into
two equal parts by the line $\mathit{il}$, one part the triangle $\mathit{lac}$, the
other the quadrilateral $\mathit{lcbg}$.

\SmallLeftNote{To this statement Leonardo adds nothing further. The proof that $k$ lies
between $a$ and $z$, and $l$ between $k$ and $z$, follows as in the first part.}

\Proposition{27}                                                          % Proposition 27
\Paragraph{}{48} \textit{``To cut off a certain fraction of a triangle    % Paragraph 48
by a straight line drawn from a given point situated outside of the
triangle}\Footnote{117}{                                                  % Footnote 117
  Some generalizations of the triangle problems in Propositions
  19, 20, 26 and   27 may be remarked. Steiner, in 1827, solved the
  problem: \textit{through a given point on a sphere to draw an arc
  of a great circle cutting two given great circles such that
  the intercepted area is equal to a given area.} (\textsc{J.~Steiner},
  ``Verwandlung und Theilung sphärischen Figuren durch Construction,''
  \textit{Crelle Jl}, \textsc{ii} (1827), pp.~56 f.
  \textit{Cf.}\ Syllabus of Townsend's course at Dublin Univ., 1846,
  in \textit{Nouvelles Annales de Mathématiques}, Sept.~1850,
  \textsc{ix}, 364; also Question 427(7) proposed by Vannson
  in \textit{Nouvelles Annales\ldots} Jan.~1858, \textsc{xvii}, 45;
  answered Aug.\ 1859, \textsc{xviii}, 335--6.)
  See also \textsc{Gudermann}, ``Über die niedere Sphärik,''
  \textit{Crelle Jl}, 1832, \textsc{viii}, 368.

  In the next year Bobillier solved, by means of planes and spheres
  only, the problem,
  \textit{to draw through a given line a plane which shall cut off
  from a given cone of revolution a required volume}.
  (\textit{Correspondance Math.\ et Physique}
  (Quetelet), \textsc{vi}$^e$ livraison,
  \textsc{iv}, 2--3, Bruxelles, 1828.)}                                    % end footnote 117
.''
[Leonardo 11, p.~121, ll.~22--23.]

Let $\mathit{abg}$ be the given triangle and $d$ the given point outside.
It is required to cut off from the triangle a certain fraction, say
one-third, by a line drawn
through $d$. Join $\mathit{ad}$, cutting $\mathit{bg}$ in $c$.
If either $\mathit{bc}$ or $\mathit{cg}$ be one-third of $\mathit{bg}$,
then the line $\mathit{ad}$ through the point $d$
cuts off one-third of the triangle $\mathit{abg}$.
But if this be not the case produce $\mathit{ab}$,
$\mathit{ag}$ to meet in $z$ and $e$ respectively
the line drawn through $d$ parallel
to $\mathit{bg}$.

\WrappedGraphic{53mm}{076}{}{r}                                           % Graphic in Proposition 27

Make
\[
  \text{area } de\centerdot gi
  = \tfrac{1}{3} \text{ area } ag\centerdot gb,
\]
and apply to the line $\mathit{gi}$ a rectangle
equal to the rectangle $\mathit{eg}\centerdot\mathit{gi}$,
but exceeded by a square; then
\[
  \mathit{eg}\centerdot\mathit{gi} = \mathit{ik}\centerdot\mathit{kg}.
\]
Draw the line $\mathit{kmd}$. I say that the triangle $\mathit{kmg}$ is one-third
of the triangle $\mathit{abg}$.

\textit{Proof:} For since
\begin{flalign*}
  &&  \text{area } eg\centerdot gi &= \text{ area } gk\centerdot  ki, && \\
  &&  eg: gk &= ki: ig. && \\
  &\text{\indent Hence}&  ek: gk &= gk: gi. &   \textsc{[v.~17]}&\\
% -----File: 077.png---Folio 65------------------------------------------------------
  &\text{\indent But }& ek:kg &= de:gm;  & \text{[\textsc{vi}.~2]} &\\
  && \therefore ed:gm &= gk:gi.  && \\
  && \therefore \text{ area } gk\centerdot gm
    &= \text{ area } de\centerdot gi.  && \\
  &\text{\indent But }
    & \text{ area } de\centerdot gi
    &= \tfrac{1}{3}\text{ area } ag\centerdot gb;  &&\\
  && \therefore \text{ area } gk\centerdot gm
    &= \tfrac{1}{3} \text{ area } ag\centerdot gb.  &&
\end{flalign*}
And since
\begin{flalign*}
  && \text{ area } gk\centerdot gm : \text{ area } ag\centerdot gb
    &= \triangle\mathit{kgm}:\triangle\mathit{agb}\footnotemark[88],  &&\\
  && \triangle\mathit{kgm} &= \tfrac{1}{3} \triangle\mathit{agb}.  &&
\end{flalign*}

In a similar manner any part of a triangle may be cut off
by a straight line drawn from a given point, on a side of the
triangle produced, or within two produced sides.

\Proposition{28}                                                          % Proposition 28
\Paragraph{}{49} \textit{``To divide into two equal parts a given figure  % Paragraph 49
bounded by an arc of a circle and by two straight lines which form a
given angle.''}
[Leonardo 57, p.~148, ll.~13--14.]

``Let $\mathit{ABC}$ be the given figure bounded by the arc $\mathit{BC}$ and by the two
lines $\mathit{AB, AC}$
which form the angle $\mathit{BAC}$.
It is required to draw a straight
line which will divide the figure
$\mathit{ABC}$ into two equal parts.

\WrappedGraphic{66mm}{077}{13}{r}                                         % Graphic in Proposition 28

``Draw the line $\mathit{BC}$ and
bisect it at $E$. Through the
point $E$ draw a line perpendicular
to $\mathit{BC}$, as $\mathit{EZ}$, and
draw the line $\mathit{AE}$. Then because
$\mathit{BE}$ is equal to $\mathit{EC}$, the
area $\mathit{BZE}$ is equal to the area
$\mathit{EZC}$, and the triangle $\mathit{ABE}$
is equal to the triangle $\mathit{AEC}$. Then the figure $\mathit{ABZE}$ will
equal the figure $\mathit{ZCAE}$. If the line $\mathit{AE}$ lie in $\mathit{EZ}$ produced,
the figure will be divided into two equal parts $\mathit{ABZE}$ and
$\mathit{CAEZ}$. But if the line $\mathit{AE}$ be not in the line $\mathit{ZE}$ produced,
join $A$ to $Z$ by a straight line and through the point $E$ draw
a line, as $\mathit{ET}$, parallel to the line $\mathit{AZ}$. Finally draw the line
% -----File: 078.png---Folio 66------------------------------------------------------
$\mathit{TZ}$. I say, that the line $\mathit{TZ}$ is that which it is required to
find, and that the figure $\mathit{ABC}$ is divided into two equal parts
$\mathit{ABZT}$ and $\mathit{ZCT}$.

``For since the two triangles $\mathit{TZA}$ and $\mathit{EZA}$ are constructed
on the same base $\mathit{AZ}$ and contained between the
same parallels $\mathit{AZ}$, $\mathit{TE}$: the triangle $\mathit{ZTA}$ is equal to the
triangle $\mathit{AEZ}$. Then, adding to each the common part
$\mathit{AZB}$, we have $\mathit{TZBA}$ equal to $\mathit{ABZE}$. But this latter figure
was half of the figure $\mathit{ABC}$; consequently the line $\mathit{ZT}$ is
the line sought and $\mathit{BZCA}$ is divided into two equal parts
$\mathit{ABZT, TZC}$, which was to be demonstrated.''

\SmallLeftNote{
Leonardo's proof is practically word for word as the above. He gives two
figures and in each he uses the \emph{Greek} succession of letters.

It is doubtless to this Proposition and the next that reference is made in
the account of Proclus [Art.~\textsc{i}].}

\Proposition{29}                                                          % Proposition 29
\Paragraph{}{50} \textit{``To draw in a given circle two parallel lines   % Paragraph 50
cutting off a certain fraction from the circle.''}
[Leonardo 51 (the case where the fraction is one-third), p.~146, ll.~37--38.]

``Let the certain fraction be one-third, and the circle be
$\mathit{ABC}$. It is required to do that which is about to be explained.

\CenteredGraphic{70mm}{078}                                               % Graphic in Proposition 29

``Construct the side of the triangle (regular) inscribed in
this circle. Let this be $\mathit{AC}$. Draw the two lines $\mathit{AD}$, $\mathit{DC}$
and draw through the point $D$ a line parallel to the line $\mathit{AC}$,
such as $\mathit{DB}$. Draw the line $\mathit{CB}$. Divide the arc $\mathit{AC}$ into
% -----File: 079.png---Folio 67------------------------------------------------------
two equal parts at the point $E$, and draw from the point $E$
the chord $\mathit{EZ}$ parallel to the line $\mathit{BC}$. Finally draw the line
$\mathit{AB}$. I say that we have two parallel lines $\mathit{EZ, CB}$ cutting
off a third of the circle $\mathit{ABC}$, viz. the figure $\mathit{ZBCE}$.

``\textit{Demonstration}. The line $\mathit{AC}$ being parallel to the line
$\mathit{DB}$, the triangle $\mathit{DAC}$ will be equal to the triangle $\mathit{BAC}$;
add to each the segment of the circle $\mathit{AEC}$; the whole figure
$\mathit{DAEC}$ will be equal to the whole figure $\mathit{BAEC}$. But the
figure $\mathit{DAEC}$ is one-third of the circle. Consequently the
figure $\mathit{BAEC}$ is also one-third of the circle. Since $\mathit{EZ}$ is
parallel to $\mathit{CB}$, the arc $\mathit{EC}$ will be equal to the arc $\mathit{BZ}$; but
$\mathit{EC}$ is equal to $\mathit{EA}$, hence $\mathit{EA}$ equals $\mathit{ZB}$. Add to these
equal parts the arc $\mathit{ECB}$; the whole arc $\mathit{AB}$ will equal the
whole arc $\mathit{EZ}$. Consequently the line $\mathit{AB}$ will be equal to
the line $\mathit{EZ}$, and the segment of the circle $\mathit{AECB}$ will be
equal to the segment of the circle $\mathit{ECBZ}$. Taking away the
common segment $\mathit{BC}$, there remains the figure $\mathit{EZBC}$ equal
to the figure $\mathit{BAEC}$. But the figure $\mathit{BAEC}$ was one-third
of the circle $\mathit{ABC}$. Then the figure $\mathit{EZBC}$ is one-third of
the circle $\mathit{ABC}$; which was to be demonstrated.

``When it is required to cut off a quarter of a circle, or a
fifth or any other definite fraction, by means of two parallel
lines, we construct in this circle the side of a square or of the
pentagon (regular) inscribed in the circle and we draw from
the centre to the extremities of this side the two straight lines
as above. (The remainder of) the construction will be
analogous to that which has gone before\Footnote{118}{                    % Footnote 118
  This problem is clearly not susceptible of solution with ruler and
  compasses, in such a case as when the ``certain fraction,''
  $\frac{1}{n}$, is one-seventh.
  In fact the only cases in which the problem is possible,
  for a fraction of this kind, is when $n$ is of the form
  \[2^p(2^{2^{s_1}}+1)(2^{2^{s_2}}+1)\dots (2^{2^{s_m}}+1),\]
  where $p$, and $s$'s (all different), are positive integers or zero,
  and $2^{2^{s_m}}+1(m=1, 2, \dots m)$ is a prime number.
  (\textit{Cf.}\ C.~F.\ Gauss,
  \textit{Disquisitiones Arithmeticae}, Lipsiae, 1801, French ed.,
  Paris, 1807, p.~489.)}                                                  % end footnote 118

\small\DPnote{continues for 60 lines}
The statement and form of discussion of this proposition are not wholly
satisfactory. For ``a certain fraction'' in the enunciation we should rather
expect ``one-third,'' as in Leonardo; while at the conclusion of the proof might
possibly occur a remark to the effect that a similar construction would apply
when the certain fraction was one-quarter [by means of \textsc{iv}.~6],
one-fifth [\textsc{iv}.~11],
one-sixth [\textsc{iv}.~15], or one-fifteenth [\textsc{iv}.~16],
but is it conceivable that Euclid
added ``or any other definite fraction''? Moreover the lack of definition of
$D$ and certain matters of form seem to further indicate that modification of
the original has taken place in its passage through Arabian channels.

% -----File: 080.png---Folio 68------------------------------------------------------
On the other hand Leonardo presents the proposition as if drawn from
the pure well of Euclid undefiled. Here is his discussion.
(I have substituted $C$ for his $b$, and $B$ for his $g$.)

``And if, by means of two parallel lines, we wish to cut off from a circle
$\mathit{ACB}$, whose centre is $D$, a given part which is one-third, draw the line $\mathit{AC}$,
the side of an equilateral triangle inscribed in the circle $\mathit{abg}$. Through the
centre $D$ draw $\mathit{DB}$ parallel to this line and join $\mathit{CB}$.
Bisect the arc $\mathit{AC}$ at
$E$ and draw $\mathit{EZ}$ parallel to $\mathit{bg}$.
I say that the figure contained between the lines $\mathit{CB}$ and $\mathit{EZ}$ and
the arcs $\mathit{EC}$ and $\mathit{BZ}$ is one-third part of the circle $\mathit{ACB}$.

``\textit{Proof:} Draw the lines $\mathit{DA}$ and $\mathit{DB}$ and $\mathit{AB}$.

``The triangles $\mathit{BAC}$ and $\mathit{DAC}$ are equal. To each add the portion $\mathit{ABE}$.
Then the figure bounded by the lines $\mathit{BA}$ and $\mathit{BC}$ and the arc $\mathit{AEC}$ is
equal to the sector $\mathit{DAEC}$ which is a third part of the circle $\mathit{ABC}$.

``Therefore the figure bounded by the lines $\mathit{BA}$ and $\mathit{BC}$ and the
arc $\mathit{AEC}$ is a third part of the circle.

``And since the lines $\mathit{CB}$ and $\mathit{EZ}$ are parallel,
the arcs $\mathit{EC}$ and $\mathit{BZ}$ are equal.
But arc $\mathit{EC}$ is equal to arc $\mathit{AE}$. Therefore arc $\mathit{AE}$ is equal to the
arc $\mathit{BZ}$.
To each add the arc $\mathit{EB}$, and then the arc $\mathit{AECB}$ will be equal to
the arc $\mathit{ECBZ}$.

``Hence the portion $\mathit{EZBC}$ of the circle is equal to the portion $\mathit{ABCE}$.
Take away the common part between the line $\mathit{CB}$ and the arc $\mathit{BC}$ and there
remains the figure, bounded by the lines $\mathit{BC}$ and $\mathit{EZ}$ and the arcs $\mathit{CE}$ and
$\mathit{BZ}$, which is the third part of the circle since it is equal to the figure bounded
by the lines $\mathit{BA}$ and $\mathit{BC}$ and the arc $\mathit{AEC}$;
quod oportebat ostendere.''

In his \ItalicGreekSnippet{metrik'a}
(\textsc{iii}.~18) Heron of Alexandria considers the problem:
\textit{To divide the area of a circle into three equal parts by two straight lines}. He
remarks that ``it is clear that the problem is not rational''; nevertheless ``on
account of its practical use'' he proceeds to give an approximate solution.
By discussion similar to that above he finds the figure $\mathit{BCEA}$, formed by
the triangle $\mathit{BCA}$ and the segment $\mathit{CEA}$, to be one-third of the circle.
Neglecting the smaller segment with chord $\mathit{BC}$, we have, that $\mathit{BA}$ cuts off
``approximately'' one-third of the circle. Similarly a second chord from $B$
might be drawn to cut off another third of the circle, and the approximate
solution be completed.
\normalsize

% -----File: 081.png---Folio 69------------------------------------------------------

\Proposition{30}                                                          % Proposition 30
\Paragraph{}{51} \textit{''To divide a given triangle into two parts by   % Paragraph 51
a line parallel to its base, such that the ratio of one of the two
parts to the other is equal to a given ratio.''}

\WrappedGraphic{53mm}{181}{12}{r}                                         % Graphic in Proposition 30

\SmallLeftNote{\DPnote{both Prop 30 and 31 are set \small. Not sure why.??}
Although Leonardo does not explicitly formulate this problem or the
next, the method to be employed is clearly
indicated in the discussion of Proposition~5 (Art.~26).

Let $\mathit{abg}$ be the triangle which is to be
divided in the given ratio $\mathit{ez:zi}$, by a line
parallel to $\mathit{bg}$. Divide $\mathit{ab}$ in $h$ such that
\[
  \mathit{ah}^2 : \mathit{ab}^2 = \mathit{ez} : \mathit{ei}\footnotemark[92].
\]

Draw $\mathit{hk}\parallel\mathit{bg}$ and meeting $\mathit{ag}$ in $k$. Then
the triangles $\mathit{ahk}$ and $\mathit{abg}$ are similar and
\[
  \triangle\mathit{ahk} : \triangle\mathit{abg} = \mathit{ah}^2: \mathit{ab}^2.
  \tag*{[\textsc{vi}.~19]}
\]
\begin{flalign*}
&\text{\indent But}&  ah^2: ab^2 &= \mathit{ez:ei}, && \\
&&  \therefore \triangle\mathit{ahk} : \triangle\mathit{abg} &= \mathit{ez:ei};&&  \\
&\text{whence}&  \triangle\mathit{ahk} : \text{quadl.}\mathit{hbgk} &= ez: zi; &[\textsc{v}.~16, 17]&
\end{flalign*}
and the triangle $\mathit{abg}$ has been divided as required.}


\Proposition{31}                                                          % Proposition 31
\Paragraph{}{52} \textit{``To divide a given triangle by lines parallel   % Paragraph 52
to its base into parts which have given ratios to one another.''}

\WrappedGraphic{50mm}{281}{10}{r}                                         % Graphic in Proposition 31

\SmallLeftNote{
Again in the manner of Proposition~5, suppose it be required to divide
the triangle $\mathit{abg}$ into three parts in the ratio\\
$\mathit{ez:zt:ti}$. Then determine the points $h$, $l$
in $\mathit{ab}$ such that
\begin{flalign*}
  && ah^2: ab^2 &= ez: ei\footnotemark[92]; && \\
  &\text{and}  & al^2: ab^2 &= et: ei.  \\
  &\text{Then}  & ah^2: al^2 &= ez: et. &&[\textsc{v}.~16, 20]
\end{flalign*}

\begin{flalign*}
  &&  \therefore\; & \triangle\mathit{ahk} : \triangle\mathit{alm} = \mathit{ez:et}, &&\\
  &\text{and} &  \therefore\triangle\mathit{ahk} & : \text{quadl.} \mathit{hlmk} = \mathit{ez:zt}.&&\\
  &\text{\indent Similarly} &&&&\\
  && \triangle\mathit{alm} & : \text{quadl.}\mathit{lbgm} = \mathit{et:ti}.  &&\\
  &\text{\indent But} & \triangle\mathit{ahk} & : \triangle\mathit{alm} = \mathit{ez:et}.  &&\\
  &&  \therefore  \triangle\mathit{ahk} & : \text{quadl.}\mathit{lbgm} = \mathit{ez:ti}.
    &[\textsc{v}.~20]&\\
  &\text{\indent Hence,}& & \triangle\mathit{ahk}: \text{quadl.}\mathit{hlmk}:
    \text{quadl.}\mathit{lbgm} = \mathit{ez:zt:ti}, &&
\end{flalign*}
and the triangle $\mathit{abg}$ has been divided into three parts in a given ratio to one
another. So also for any number of parts which have given ratios to one
another.}
% -----File: 082.png---Folio 70------------------------------------------------------

\Proposition{32}                                                          % Proposition 32
\Paragraph{}{53} \textit{``To divide a given trapezium by a line parallel % Paragraph 53
to its base, into two parts such that the ratio of one of these parts
to the other is equal to a given ratio.''}
[Leonardo 29, p.~131, ll.~41--42.]

Let $\mathit{abgd}$ be the trapezium which is to be divided in the ratio $\mathit{ez:zi}$
by a line parallel to
the base. Produce the sides $\mathit{ba}$,
$\mathit{gd}$ through $a$ and $d$ to meet in $t$.

\WrappedGraphic{53mm}{082}{16}{r}                                         % First Graphic in Proposition 32

\begin{DPalign*}
  \lintertext{\indent Make} tl^2 : at^2 &= zi : ez\footnotemark[92], \\
  \lintertext{and} ht^2 : (bt^2 + tl^2) &= ez : ei.
\end{DPalign*}

Through $l$, $h$, draw $\mathit{lm}$, $\mathit{hk}$
parallel to $\mathit{bg}$ and $\mathit{ad}$. Then I say
that the quadrilateral $\mathit{ag}$ is divided
in the given ratio, $\mathit{ez:zi}$, by the
line $\mathit{hk}$.

\textit{Proof:} For since the triangles
$\mathit{tlm, tad}$ are similar
\[
  tl^2 : at^2 = \triangle\mathit{tlm} : \triangle\mathit{tad};
\]

\begin{flalign*}
  &\text{but} & tl^2 : at^2 &= zi : ez; && \\
  && \therefore zi : ez     &= \triangle\mathit{tlm} : \triangle\mathit{tad}. && \\
  &\text{Whence} & ei : ez  &= (\triangle\mathit{tlm} + \triangle\mathit{tad}) : \triangle\mathit{tad},     & \textsc{[v.~18]}& \\
  &\text{or} & ez : ei      &= \triangle\mathit{tad} : (\triangle\mathit{tlm} + \triangle\mathit{tad}).     & \textsc{[v.~16]}&
\end{flalign*}

But by construction
\begin{flalign*}
  && \mathit{ez:ei} &= \mathit{ht}^2 : (\mathit{bt}^2 + \mathit{tl}^2), && \\
  &\text{and} & ht^2 : (bt^2 + tl^2) &=\triangle\mathit{thk} : (\triangle\mathit{tbg} +
    \triangle\mathit{tlm}). & \textsc{[vi.~19]}& \\
  && \therefore \mathit{ez:ei} &= \triangle\mathit{thk} : (\triangle\mathit{tbg} +
      \triangle\mathit{tlm}). &&\\
  &\text{\indent But}&
    \triangle\mathit{thk}&=\triangle\mathit{tad}+\text{quadl.}\mathit{ak}. &&
\end{flalign*}

Similarly
\begin{gather*}
  \triangle\mathit{tbg} + \triangle\mathit{tlm} = \text{quadl.}\mathit{ag} + \triangle\mathit{tad} +
      \triangle\mathit{tlm}.\\
  \therefore ez : ei = (\text{quadl.}\mathit{ak} + \triangle\mathit{tad}) :
    (\text{quadl.}\mathit{ag} + \triangle\mathit{tad} + \triangle\mathit{tlm}).
\end{gather*}
% -----File: 083.png---Folio 71------------------------------------------------------
\begin{flalign*}
  & \text{\indent But }
    & \mathit{ez:ei} &= \triangle\mathit{tad}: (\triangle\mathit{tad} + \triangle\mathit{tlm});  &&\\
  && \therefore \mathit{ez:ei} &= \text{quadl.}\mathit{ak}:\text{quadl.}\mathit{ag};
    &  \text{[\textsc{v}.~11,~19]}&\\
  & \text{whence} & \mathit{ez:zi} &= \text{quadl.}\mathit{ak}:\text{quadl.}\mathit{hg}.  &&
\end{flalign*}
And the trapezium has been divided in the given ratio.

\SmallLeftNote{Then follows a numerical example and this alternative construction and proof:}

Draw $\mathit{mls}$ such that,
\[
  \mathit{ms:ls} = \mathit{tb}^2:\mathit{ta}^2,\footnotemark[92]
\]
and divide $\mathit{ml}$ in $n$, such that $\mathit{ln}$ is to $\mathit{nm}$ in the given proportion.

\CenteredGraphic{70mm}{083}                                               % Second Graphic in Proposition 32

In $\mathit{tb}$ determine $h$ such that
\[
  \mathit{th}^2: \mathit{tb}^2 = \mathit{ns:sm}.
\]

Draw $\mathit{hk}\parallel\mathit{bg}$. Then,
\[
  \text{quadl.}\mathit{ak}:\text{quadl.}\mathit{hg} = \mathit{ln:nm}.
\]
\begin{flalign*}
  & \text{\indent \textit{Proof:} For, }
    & tb^2: ta^2 &= \triangle\mathit{tbg}: \triangle\mathit{tad};
    & \text{[\textsc{vi}.~19]}  & \\
  & \text{and} & \mathit{ms:ls} &= \mathit{tb}^2: \mathit{ta}^2.  &&\\
  &&  \therefore\mathit{ms:ls} &= \triangle\mathit{tbg}: \triangle\mathit{tad}.  &
    \ldots\ldots\ldots\text{[1]} &\\
  & \text{Again, since} & \mathit{tb}^2: \mathit{th}^2 &= \mathit{ms:sn},  &&\\
  & \text{and} & \triangle\mathit{tbg}: \triangle\mathit{thk} &= tb^2: th^2,  &&\\
  &&  \therefore\mathit{ms:ns} &= \triangle\mathit{tbg}: \triangle\mathit{thk};  &
    \ldots\ldots\ldots\text{[2]} &\\
  &&  \therefore\mathit{sm:nm} &= \triangle\mathit{tbg}: \text{quadl.}\mathit{hg}.
    & \text{[\textsc{v}.~16,~21] \ldots\ldots[3]}&\\
% -----File: 084.png---Folio 72------------------------------------------------------
  &\text{But }
    & \mathit{sm:ls} &= \triangle\mathit{tbg}: \triangle\mathit{tad}\footnotemark[1181],  &&\\
  &\text{or} &  \mathit{ms}: \triangle\mathit{tbg} &= \mathit{ls}: \triangle\mathit{tad},  &&\\
  &\text{while } & \mathit{ms}: \triangle\mathit{tbg} &= \mathit{ns}: \triangle\mathit{thk};
    &\text{[from [2]]} &\\
  &&  \therefore\mathit{ls : ns} &= \triangle\mathit{tad}:\triangle\mathit{thk}.  &
    \ldots\ldots\ldots\text{[4]} &\\
  &\text{\indent From [3]}
    & \mathit{ms}:\triangle\mathit{tbg} &=\mathit{nm}: \text{quadl.}\mathit{hg}.  &&\\
  &\text{\indent But from [4]}
    & \mathit{sl : ln} &= \triangle\mathit{tad}:\text{quadl.}\mathit{ak},
    &\text{[\textsc{v}.~16, 21]} & \\
%% The next four lines were originally set in a smaller font.
  &&  \therefore\mathit{sl}:\triangle\mathit{tad} &= \mathit{ln}:\text{quadl.}\mathit{ak}.  &&\\
  &\text{\indent But from [1]}
  & \mathit{ms}:\triangle\mathit{tbg} &= \mathit{sl}:\triangle\mathit{tad},  &&\\
  &&  \therefore\mathit{ms}:\triangle\mathit{tbg} &= \mathit{ln}: \text{quadl.}\mathit{ak};  && \\
  &&  \therefore\mathit{mn}: \text{quadl.}\mathit{hg} &= \mathit{ln}:\text{quadl.}\mathit{ak};  &&\\
  &&  \therefore\mathit{ln : nm} &= \text{quadl.}\mathit{ak}: \text{quadl.}\mathit{hg}.  &&
\end{flalign*}
\Footnotetext{1181}{                                                      % Footnote 1181 (was 118a)
  Such mixed ratios as these (ratios of lines to areas),
  and others of like kind which follow in this proof,
  are very un-Greek in their formation. This is sufficient to stamp
  the second proof as of origin other than Greek. The first proof, on
  the other hand, is distinctly Euclidean in character.}                  % end footnote 1181

Hence the quadrilateral $\mathit{ag}$ is divided by the line $\mathit{hk}$,
parallel to the base $\mathit{bg}$, in the given proportion as the
number $\mathit{ln}$ is to the number $\mathit{nm}$. Which was to be done.

\SmallLeftNote{Then follows a numerical example.}

\Proposition{33}                                                          % Proposition 33
\Paragraph{}{54} \textit{``To divide a given trapezium, by lines parallel % Paragraph 54
to its base, into parts which have given ratios to one another.''}
[Leonardo 35, p.~137, ll.~6--7.]

Let $\mathit{abgd}$ be the given trapezium and [let $\mathit{ez:zi:it}$ denote
\WrappedGraphic{65mm}{084}{}{r}                                           % Graphic in Proposition 33
the ratios of the three parts
into which the trapezium is to
be divided by lines parallel to
the base $\mathit{bg}$]. Let $\mathit{ba,gd}$ produced
meet in $k$ and find $l$ such
that
\[
  bk^2:ak^2 = \mathit{tl:el}.
\]
Then determine $m$ and $n$ such
that
\begin{flalign*}
  &            & \mathit{bk}^2: \mathit{km}^2 &= \mathit{tl:lz},  &&\\
  &\text{and} & \mathit{bk}^2: \mathit{kn}^2 &= \mathit{tl:il}.  &&
\end{flalign*}

% -----File: 085.png---Folio 73------------------------------------------------------

Through $\mathit{m,n}$ draw lines $\mathit{mo,np}$ parallel to $\mathit{bg}$. In the
same manner as above
\begin{flalign*}
  &          & \text{quadl.}\mathit{ao}:\text{quadl.}\mathit{mp} &=\mathit{ez:zi};  &&\\
  &\text{and}& \text{quadl.}\mathit{mp}:\text{quadl.}\mathit{ng} &=\mathit{zi:it}.  &&
\end{flalign*}

\small\DPnote{continues for 40 lines}
Then follows a numerical example in which the line $\mathit{kfhrs}$, perpendicular
to $\mathit{bg}$, is introduced into the figure.

Here is a proof of the Proposition:

By construction, \textsc{v}.~16 and \textsc{vi}.~19,
\begin{flalign*}
  &                            &&\triangle\mathit{kbg}:\triangle\mathit{kad}=\mathit{tl:el}.
      &\ldots\ldots\ldots\ldots\ldots\ldots\text{[1]}& \\
  &\text{\indent So also}      &&\triangle\mathit{kbg}:\triangle\mathit{kmo}=\mathit{tl:lz},
      &\ldots\ldots\ldots\ldots\ldots\ldots\text{[2]}& \\
  &\text{and }                 &&\triangle\mathit{kbg}:\triangle\mathit{knp}=\mathit{tl:il}
      &\ldots\ldots\ldots\ldots\ldots\ldots\text{[3]}&\\
  &\text{\indent From [1] }    &&\triangle\mathit{kad}:\triangle\mathit{kbg}=\mathit{el:tl}.  &&\\
  &\text{\indent But from [2] }&&\triangle\mathit{kbg}:\triangle\mathit{kmo}=\mathit{tl:lz};  &&\\
  &\text{hence, by [\textsc{v}.~20],}&&\triangle\mathit{kad}:\triangle\mathit{kmo}=\mathit{el:lz},
        &\ldots\ldots\ldots\ldots\ldots\ldots\text{[4]}&\\
  &\text{or alternately}&  &\triangle\mathit{kmo}:\triangle\mathit{kad}=\mathit{lz:el}.  &&\\
  &\text{\indent Hence,
    \textit{separando},}&&\text{quadl.}\mathit{ao}:\triangle\mathit{kad}=\mathit{ez:el}.
    &\ldots\ldots\ldots\ldots\ldots\ldots\text{[5]}&\\
  &\rlap{\text{\indent So also from [2] and [3]}}&  &  &&\\
  &&  &\triangle\mathit{kmo}:\triangle\mathit{knp}=\mathit{lz:il};  &&\\
  &\text{and}&  &\triangle\mathit{kmo}:\text{quadl.}\mathit{mp}=\mathit{lz:iz}.  &&\\
  &\text{\indent But from [4]}&&\triangle\mathit{kad}:\triangle\mathit{kmo}=\mathit{el:lz};  &&\\
  &\text{therefore, by [\textsc{v}.~20], }
    &&\triangle\mathit{kad}:\text{quadl.}\mathit{mp}=\mathit{el:iz}
    &\ldots\ldots\ldots\ldots\ldots\ldots\text{[6]}&\\
%%%\end{flalign*}
%%%\begin{flalign*}
  &\rlap{\text{\indent Hence from [5], by [\textsc{v}.~20],}} &&&&\\
  && &\text{quadl.}\mathit{ao}:\text{quadl.}\mathit{mp}=\mathit{ez:zi}. &&\\
  &\text{\indent Again, from [3],}&&\text{quadl.}\mathit{ng}:\triangle\mathit{kbg}=\mathit{ti:tl};  &&\\
  &\text{and since from [1], }&&\quad\triangle\mathit{kbg}:\triangle\mathit{kad}=\mathit{tl:el},  &&\\
  &\text{we have}&&\text{quadl.}\mathit{ng}:\triangle\mathit{kad}=\mathit{it:el}.  &&\\
  &\intertext{\indent Hence from [6], by [\textsc{v}.~20], we get}   \\
  &                     &&\text{quadl.}\mathit{ng}:\text{quadl.}\mathit{mp}=\mathit{it:zi},  &&\\
  &\text{or alternately}&&\text{quadl.}\mathit{mp}:\text{quadl.}\mathit{ng}=\mathit{zi:it}.  &&
\end{flalign*}

And since $\text{quadl.}\mathit{ao}:\text{quadl.}\mathit{mp}=\mathit{ez:zi}$,
the trapezium $\mathit{ag}$ has been divided
by lines parallel to the base $\mathit{ag}$, into three parts which are in the required
ratios to one another.      \hfill           \textsc{q.~e.~f.}\qquad
\normalsize
% -----File: 086.png---Folio 74------------------------------------------------------

\Proposition{34}                                                          % Proposition 34
\Paragraph{}{55} \textit{``To divide a given quadrilateral, by a line     % Paragraph 55
drawn from a given vertex of the quadrilateral, into two parts such
that the ratio of one of these parts to the other is equal to a
given ratio.''}
[Leonardo 40, p.~140, ll.~36--37.]

Let $\mathit{abcd}$ be the given quadrilateral, and $\mathit{ez:zi}$ the given ratio.
It is required to draw from the
angle $d$ a line to divide the
quadrilateral in the ratio $\mathit{ez:zi}$.
\WrappedGraphic{70mm}{086}{}{r}                                           % Graphic in Proposition 34

Draw the diagonal $\mathit{ac}$ and
on it find $t$ such that
\[
  \mathit{ct:at}=\mathit{ez:zi}.
\]

Draw the diagonal $\mathit{bd}$. Then
if $\mathit{bd}$ pass through $t$ the quadrilateral
is divided as required,
in the ratio $\mathit{ez:zi}$.

For,
\begin{flalign*}
  &&  \triangle\mathit{dct}:\triangle\mathit{dta} &= \mathit{ct:ta},  &&\\
  &&                               &= \triangle\mathit{cbt}:\triangle\mathit{tba}.  &&\\
  &&             \therefore\mathit{ct:ta} &= \triangle\mathit{dcb}:\triangle\mathit{abd}.
    &&[\textsc{v}.~18]\\
  &\text{\indent But} &\mathit{ct:ta}&= \mathit{ez:zi};  &&\\
  &&             \therefore\mathit{ez:zi} &= \triangle\mathit{bdc}:\triangle\mathit{bda}; &&
\end{flalign*}
and the quadrilateral $\mathit{ac}$ is divided, by a line drawn from a
given angle, in a given ratio.

But if $\mathit{bd}$ do not pass through $t$, it will cut $\mathit{ca}$ either
between $c$ and $t$ or between $t$ and $a$. Consider first when
$\mathit{bd}$ cuts $\mathit{ct}$. Join $\mathit{bt}$ and $\mathit{td}$. Then,
\[
  \text{quadl.}\mathit{tbcd}:\text{quadl.}\mathit{tbad} = \mathit{ct:ta}=\mathit{ez:zi}.
\]

Draw $\mathit{tk}\parallel\mathit{bd}$ and join $\mathit{dk}$. Then
\begin{DPalign*}
                             \text{quadl.}\mathit{kbcd} &= \text{quadl.}\mathit{tbcd};  \\
  \therefore\text{quadl.}\mathit{kbcd}:\triangle\mathit{dak} &=\mathit{ez:zi},
\end{DPalign*}
and the line $\mathit{dk}$ has been drawn as required.

% -----File: 087.png---Folio 75------------------------------------------------------

If the diagonal $\mathit{bd}$ cut $\mathit{at}$, through $t$ draw $\mathit{tl}$
parallel to the
diagonal $\mathit{bd}$. Join $\mathit{dl}$. Then as before,
\[
  \mathit{ct:ta}=\mathit{ez:zi}=\triangle\mathit{dcl}:\text{quadl.}\mathit{abld}.
\]

\CenteredGraphic{68mm}{187}                                               % Graphic in Proposition 34

\SmallLeftNote{
Hence in every case the quadrilateral has been divided as required by a
line drawn from $d$. Similarly for any other vertex of the quadrilateral.}

\Proposition{35}                                                          % Proposition 35
\Paragraph{}{56} \textit{``To divide a given quadrilateral by lines       % Paragraph 56
drawn from a given vertex of the quadrilateral into parts which
are in given ratios to one another.''}

\SmallLeftNote{
Although Leonardo does not explicitly formulate this problem, the
method he would have followed is clear from his discussion of the last
Proposition. Let $\mathit{abcd}$ be the quadrilateral to be divided, by lines drawn
from $d$, into three parts in the ratios to one another of $\mathit{ez:zi:it}$.

\CenteredGraphic{72mm}{287}                                               % Graphic in Proposition 35

Divide $\mathit{ca}$ at points $\mathit{r,t}$ so that
\[
  \mathit{cr:rt:ta}=\mathit{ez:zi:it}.
\]

Through $\mathit{r,t}$ draw lines parallel to $\mathit{bd}$, and meeting $\mathit{bc}$
(or $\mathit{ab}$) in $l$ and
$\mathit{ab}$ (or $\mathit{bc}$) in $m$.

Then as above $\mathit{dl}$, $\mathit{dm}$ divide the quadrilateral as required.

We may proceed in a similar manner to divide the quadrilateral $\mathit{abcd}$, by
lines drawn from the angular point $d$, into any number of parts in given ratios
to one another.}

% -----File: 088.png---Folio 76------------------------------------------------------

\Proposition{36}                                                          % Proposition 36
\Paragraph{}{57} \textit{``Having resolved those problems which           % Paragraph 57
have gone before, we are in a position to divide a given quadrilateral
in a given ratio or in given ratios by a line or by lines drawn
from a given point situated on one of the sides of the quadrilateral,
due regard being paid to the conditions mentioned above.''}

\small\DPnote{continues for 92 lines}
This problem, also, is not formulated by Leonardo; but from his discussion
of Euclid's Propositions 16, 17 and of his own 41, the methods of construction
which Euclid might have employed are clearly somewhat as follows.

Let $\mathit{abcd}$ be the given quadrilateral and $g$ the given point.

\CenteredGraphic{85mm}{188}                                               % Graphic in Proposition 36

(1) Let it be required to divide $\mathit{abcd}$ into two parts in the ratio $\mathit{ez:zi}$ by
a line drawn through a point $g$ in the side $\mathit{ad}$.

Draw $\mathit{dl}$ such that
 $\triangle\mathit{cld}:\text{quadl.}\mathit{lbad}=\mathit{ez:zi}$.  \hfill [Prop.~34] \qquad\newline
Join $\mathit{gl}$. If $\mathit{gl}\parallel\mathit{dc}$, join $\mathit{gc}$,
then this line divides the quadrilateral as required.

If $\mathit{gl}$ be not parallel to $\mathit{dc}$ draw $\mathit{dh}\parallel\mathit{gl}$,
and meeting $\mathit{bc}$ in $h$. Join $\mathit{gh}$.
Then $\mathit{gh}$ divides the quadrilateral as required.

If $\mathit{dh}$ fell outside the quadrilateral
draw $\mathit{ll'}\parallel\mathit{cd}$ (not indicated in the figure)
to meet $\mathit{ad}$ in $\mathit{l'}$.
Draw $\mathit{l'z'}\parallel\mathit{gc}$ to meet $\mathit{dc}$ in $\mathit{z'}$.
Then $\mathit{gz'}$ is the line required.

The above reasoning is on the assumption that $\mathit{dl}$ meets $\mathit{bc}$ in $l$. Suppose
now it meet $\mathit{ab}$ in $l$. Join $\mathit{bd}$ and
draw $\mathit{bk}$ such that
\[
\text{quadl.}\mathit{bcdk}:\triangle\mathit{kab}=\mathit{ez:zi}.
\]

\WrappedGraphic{85mm}{288}{}{r}                                           % Graphic in Proposition 36

If $k$ do not coincide with $g$ there
are two cases to consider according
as $k$ is between $g$ and $d$ or between
$g$ and $a$. Consider the former case.

Through $k$ draw $\mathit{km}$ parallel to $\mathit{gb}$
and meeting $\mathit{bc}$ in $m$. Join $\mathit{gm}$. Then
this is the required line dividing the
quadrilateral $\mathit{ac}$ in such a way that
\[
\text{quadl.}\mathit{abmg}:\text{quadl.}\mathit{gmcd}=\mathit{ez:zi}.
\]
% -----File: 089.png---Folio 77------------------------------------------------------

Similarly if $k$ were between $g$ and $a$.

(2) Let it be required to divide $\mathit{abcd}$ into, say, three parts in the ratios
$\mathit{ez:zi:it}$, by lines through any point $g$ in the side $\mathit{ad}$ (first figure).

Draw $\mathit{dl,dm}$ dividing the quadrilateral $\mathit{ac}$ into three parts such that
\begin{equation*}
\triangle\mathit{amd}:\text{quadl.}\mathit{dmbl}:\triangle\mathit{dlc}=\mathit{ez:zi:it}.
\end{equation*}

There are various cases to consider according as $l$ and $m$ are both
on $\mathit{bc}$, both on $\mathit{ab}$, or one on $\mathit{ab}$ and one on $\mathit{bc}$.
The method will be obvious from working
out one case, say the last.

Join $\mathit{gc,gl}$. If $\mathit{gl}$ be parallel to $\mathit{cd}$,
$\mathit{gc}$ cuts off the triangle $\mathit{gdc}$ such that
\begin{flalign*}
  &&  \triangle\mathit{gdc}:\text{quadl.}\mathit{abcg}&=\mathit{it:ei \qquad(=ez+zi)}.
    &\text{ [\textsc{v}.~24]}\qquad&
\end{flalign*}

If $\mathit{gl}$ be not parallel to $\mathit{dc}$,
draw $\mathit{dh}$ parallel to $\mathit{gl}$ and meeting $\mathit{bc}$ in $h$;
then $\mathit{gh}$ divides the quadrilateral in such a way that
\[
  \text{quadl.}\mathit{gdch}: \text{quadl.}\mathit{ghba}=\mathit{it:ei}.
\]

Then apply Proposition 34 to draw from $g$ a line to divide the quadrilateral
$\mathit{abhg}$ in the ratio of $\mathit{ez:zi}$.

Hence from $g$ are drawn two lines which divide the quadrilateral $\mathit{abcd}$
into three parts whose areas are in the ratios of $\mathit{ez:zi:it}$.

The case when $\mathit{dh}$ meets $\mathit{bc}$ produced may be considered as above.

We could proceed in a similar manner if the quadrilateral $\mathit{abcd}$ were to be
divided by lines drawn from $g$, into a greater number of parts in given ratios.

The enunciation of this proposition is a manifest corruption of what Euclid
may have given. Such clauses as those at the beginning and end he would
only have included in the discussion of the construction and proof.
\normalsize

\bigskip
After the enunciation of Proposition 36, Woepcke's translation
of the Arabian MS. concludes as follows:

``End of the treatise. We have confined ourselves to
giving the enunciations without the demonstrations, because
the demonstrations are easy.''
\

% -----File: 090.png---Folio 78------------------------------------------------------


\Chapter{IV}                                                             % Chapter IV
\phantomsection\pdfbookmark[0]{Appendix}
                              {Appendix}
\flushbottom\DPnote{Chapter III was set raggedbottom; return to flushbottom.}
\thispagestyle{empty}

\label{appendix}
\label{App:appendix}
\medskip

\LargeCenteredNote{APPENDIX}
\small
\medskip

In the earlier pages I have referred to works on Divisions of Figures
written before 1500. Several of these were not published till later; for
example, that of Muhammed Bagdedinus in 1570, of Leonardo Pisano in
1862 and the second edition of Luca Paciuolo's ``Summa''
in 1523\Footnote{119}{                                                    % Footnote 119
  A synopsis of the portion of the work on divisions of figures is
  given on pages 106 and 275--284 of
  \textit{Scritti inediti del P.~D.\ Pietro Cossali\ldots}pubblicati
  da Baldassarre Boncompagni,
  Roma, 1857. \textit{Cf}.\ note 61.}.                                    % end footnote 119
It has been remarked that Fra Luca's treatment of the subject was based
entirely upon that of Leonardo. But, on account of priority in publication,
to Paciuolo undoubtedly belongs the credit of popularizing the problems on
Divisions of Figures.

While few publications treat of the subject in the early part, their number
increases in the latter part, of the sixteenth century. In succeeding centuries
the tale of titles is enormous and no useful purpose would be served by the
publication here of an even approximately complete list. It would seem,
however, as if the subject matter were of sufficient interest to warrant, as
completion of the history of the problems, a selection of such references in
this period, (1) to standard or popular works, (2) to the writings of eminent
scientists like Tartaglia, Huygens, Newton, Kepler and Euler; (3) to special
articles, pamphlets or books which treat parts of the subject; (4) to discussions
of division problems requiring other than Euclidean methods for
their solution.

No account is taken of the extensive literature on the division of the
circumference of a circle, from which corresponding divisions of its area
readily flow. Considerations along this line may be found in:
\textsc{P.~Bachmann},
\textit{Die Lehre von der Kreisteilung und ihre Beziehungen zur Zahlentheorie},
Leipzig, 1872, 12 + 300 pp.; and in \textsc{A.~Mitzscherling},
\textit{Das Problem der Kreisteilung},
Leipzig und Berlin, 1913, 6 + 214 pp.

Except for about a dozen titles, all the books or papers mentioned have
been personally examined. In many cases it will be found that only a single
problem (often Euclid's Propositions 19, 20, 26 or 27) is treated in the place
referred to.

Some titles in note 111 may also be regarded as forming a supplement to
this list.

\noindent

\begin{Description}
\AppItem{1539}{1539}
\textsc{W.~Schmid}.
\textit{Das Erst} [Zweit, Dritt und Viert] \textit{Buch der Geometria}.
Nürnberg.

{\footnotesize ``Dritter Theil, von mancherley Art der Flächen, wie dieselben gemacht und
ausgetheilt werden, auch wie eine Fläche in die andern für sich selbst, oder gegen
einer andern in vorgenommener Proporz, geschätzt, verändert mag werden. Theilungen
und Zeichnungen von Winkeln, Figuren, ordentlichen Vielecken, die letzten, wie man
leicht denken kann, nicht alle geometrisch richtig. Verwandlung von Figuren.''}

\AppItem{1547}{1547a}
\textsc{L.~Ferrari}. A ``Cartello'' which begins:
``Messer N.~Tartaglia, \DPtypo{gia}{già}
\DPnote{[** cf. http://edit16.iccu.sbn.it/scripts/iccu_ext.dll?fn=10&i=40964]}
otto giorni, cioè alli 16 di Maggio, in risposta della mia replica io riceuetti
la uostra tartagliata, etc.'' [Milan.]

{\footnotesize Dated June 1, 1547; a challenge to a mathematical disputation from L.~F.\ to
N.~Tartaglia.}

% -----File: 091.png---Folio 79------------------------------------------------------

\AppItem{1547}{1547b}
\textsc{N.~Tartaglia.}
\textit{Terza Risposta data da
N.~Tartalea\ldots al eccellente
M.~H. Cardano\ldots et al eccellente Messer L. Ferraro\ldots con la resolutione,
ouer risposta de 31 quesiti, ouer quistioni da quelli a lui proposti.}
[Venice, 1547.]

\footnotesize
Dated July 9, 1547. For the discussion between Ferrari (1522--1565) and Tartaglia
(1500--1557) 6 ``cartelli'' by Ferrari and 6 ``Risposte'' by Tartaglia were published at
Milan, Venice and Brescia in 1547--48\Footnote{120}{                      % Footnote 120
\textit{Cf.}\ \textsc{Cantor}, \textit{Vorlesungen über Gesch.\ d.\ Math.}
Bd \textsc{ii}, 2te Aufl., 1900, pp.~490--491,
where the exact dates are given.}.                                        % end footnote 120
They contained the problems and their
solutions. These publications are of excessive rarity. Only about a dozen copies
(which are in the British Museum and Italian Libraries) are known to exist. But they
have been reprinted in:
\textit{I sei cartelli di matematica disfida primamente intorno alla
generale risoluzione delle equazioni cubiche, di Ludovico Ferrari,
coi sei contro-cartelli in
risposta di Nicolo Tartaglia\ldots comprendenti le soluzioni de' quesiti dall' una e dall' altra
parte proposti\ldots Raccolti, autografati e pubblicati da Enrico Giordani.}\ldots Milano 1876.

On pages 6--7 of the \textsc{iii}$^\circ$ cartello (Giordani's edition pp.~94--95), Questions 5 and
14, proposed by Ferrari, are:---

``5. To bisect, by a straight line, an equilateral, but not equiangular, heptagon.''

``14. Through a point without a triangle to draw a line which will cut off a third.''

On pages 12 and 20 of the \textsc{iii}$^\circ$ Risposta Tartaglia gives the solutions and assigns
due credit to the treatment of problems on the Division of Figures by Luca Paciuolo.
The general subject was treated much more at length by Tartaglia in a part of his
``General trattato'' published in 1560.
\small

\AppItem {1560}{1560}
\textsc{N.~Tartaglia.}
\textit{La quinta parte del general trattato de' numeri et misure.}
Venetia.

{\footnotesize On folio 6 \textit{recto} we have a section entitled ``Del modo di
saper dividere una \DPtypo{figure}{figura}
cioè pigliar, over formar una parte di quella in forma propria.'' The division of figures
is treated on folios 23
\textit{verso}--44 \textit{recto}
(23--32, triangles; 32--34, parallelograms; 34--44,
quadrilateral, pentagon, hexagon, heptagon, circle without the Euclid-Proclus case).

\textit{Cf.}\ the synopsis in
\textit{Scritti inediti del P.~D.\ Pietro Cossali chierico regolare teatino}
pubblicati da Baldassarre Boncompagni, Roma, 1857, pp.~299--300.}

\AppItem{1574}{1574a}
\textsc{J.~Gutman.}
\textit{Feldmessung gewiss, richtig und kurz gestellt.} Heidelberg.

\AppItem{1574}{1574b}
\textsc{E.~Reinhold.}
\textit{Gründlicher und wahrer Bericht vom Feldmessen,
samt allem, was dem anhängig, darinn alle die Irrthum, so bis daher im
Messen fürgeloffen, entdeckt werden. Dessgleichen vom Markscheiden,
kurzer und gründlicher Unterricht.} Erffurdt.

{\footnotesize ``Der dritte Theil von Theilung der Aecker. Theilungen aller Figuren, auch des
Kreises mit Exemplen und Tafeln erläutert.''}

\AppItem{1585}{1585}
\textsc{G.~B.~Benedetti.}
\textit{Diversarum speculationum mathematicarum et
physicarum liber.} Taurini.

{\footnotesize Pages 304--307.}

\AppItem{1604}{1604}
\textsc{C.~Clavius.}
\textit{Geometria Practica.} Romae.

{\footnotesize Pages 276--297.}

\AppItem{1609}{1609}
\textsc{J.~Kepler.}
\textit{Astronomia nova\ldots commentariis de motibus stellæ
Martis}\ldots [Pragae].

\footnotesize
``Kepler's Problem'' occurs on p.~300 of this work
(\textit{Opera Kepleri} ed.~Frisch,
\textsc{iii}, 401).
It is: ``To divide the area of a semicircle in a certain ratio by a straight
line drawn through a given point on the diameter or on the diameter produced.''
(\textit{Cf.}\
\textsc{A.~G.~Kästner}, \textit{Geschichte der Math}.
 \textsc{iv}, 256, Göttingen, 1800;
\textsc{M.~Cantor}, \textit{Vorlesungen}
etc., \textsc{ii}, 708, Leipzig, 1900). Kepler was led to this problem in his
theory of the path of the planets. It has been attacked by many mathematicians,
notably by Wallis, Hermann, Cassini, D.~Gregory, T.~Simpson, Clairaut, Lagrange,
% -----File: 092.png---Folio 80------------------------------------------------------
Bossut, and Laplace. (\textit{Cf}.\ \textsc{G.~S.~Klügel},
\textit{Mathematisches Wörterbuch}\ldots Erste Abtheilung,
Dritter Theil, Leipzig, 1808; Article, ``Kepler's Aufgabe.'' See also
\textsc{C.~Hutton}, \textit{Philosophical and Mathematical Dictionary}.
New edition, London, 1815, 1, 703.)
\small

\AppItem{1612}{1612}
\textsc{Sybrandt Cardinael.}
\textit{Hondert geometrische questien met hare solutien}. Amsterdam.

{\footnotesize
This work is also to be found at the end of \textsc{Johan Sems} ende \textsc{Ian Dou}
\textit{Practijck des landmetens}. Amsterdam, 1616. Another edition: Tractatus geometricus.
Darinnen hundert schöne \ldots Questien [übersetzt] durch Sebastianum Curtium.
Amsterdam, 1617; Questions 78, 90--93.

With these problems Huygens (1629--1695) busied himself when about 17 or 18
years of age. \textit{Cf.\ Oeuvres complètes de Chr.\ Huygens}, Amsterdam, XI, 24 and 29, 1908.

I have elsewhere (\textit{Nieuw Archief}, 1914) shown that Sybrandt Cardinael's work
was translated into English, rearranged and published as an original work by
Thomas Rudd (1584?-1656): \textit{A hundred geometrical Questions with their solutions
and demonstrations}. London, M.DC.L.}

\AppItem{1615}{1615}
\textsc{Ludolph van Ceulen.}
\textit{Fundamenta arithmetica et geometrica cum
eorundem usu \ldots e vernaculo in Latinum translata a W.~S}[\textit{nellio}],
\textit{R.~F.} Lugduni Batavorum.

{\footnotesize Contains several problems on Change, and Division, of Figures.}

\AppItem{1616}{1616}
\textsc{J.~Speidell.}
\textit{A geometricall Extraction or a compendiovs collection of
the chiefe and choyse Problemes, collected out of the best, and latest writers.
Wherevnto is added about 30 Problemes of the Authors Invention, being for
the most part, performed by a better and briefer way, than by any former
writer}. London.

{\footnotesize Another edition, 1617; second edition ``corrected and enlarged,'' London, 1657;
``Now followeth [pp.~84--125] a compleat Instruction of the diuision of all right lined
figures\ldots. Very pleasant and full of delight in practise: Also, most profitable to all
surveighers, or others that are desirous to make any Inclosure.''}

\AppItem{1619}{1619}
\textsc{A.~Anderson.}
\textit{Exercitationum mathematicarum Decas prima. Continens,
Questionum aliquot, quae nobilissimorum tum hujus tum veteris
aevi, Mathematicorum ingenia exercuere, Enodationem}. Parisiis.

{\footnotesize Problems in division of a triangle, with reference to Clavius (1604).
\textit{Cf.\ The Ladies' Diary}, London, 1840, pp.~55--56.}

\AppItem{1645}{1645}
\textsc{C.~Huygens.}
\textit{Oeuvres Complètes}, \textsc{xi}, 1908, pp.~26--27; 219--225.

{\footnotesize Solution of ``Datum triangulum, ex puncto in latere dato, bifariam secare'' (1645);
two solutions of ``Triang.\ ABC, sectus utcumque lineâ DE, dividendus est aliâ lineâ,
FG, ita ut utraque pars DBE et ADEC bifariam dividatur'' (1650--1668). See also
note under 1687---J.~Bernoulli.}

\AppItem{1657}{1657}
\textsc{F.~van Schooten.}
\textit{Exercitationvm mathematicarum liber primus
continens propositionum arithmeticarvm et geometricarvm centuriam}.
Lugd.\ Batav.

{\footnotesize Prop.~L, pp.~107--110. Dutch edition, Amsterdam, 1659. pp.~107--110. Concerning
a Schooten MS. of 1645--6, used by Huygens and of interest in this connection,
\textit{cf.}\ \textsc{C.~Huygens,} \textit{Oeuvres Complètes}, tome \textsc{xi}, 1908, p.~13~ff.}

\AppItem{1667}{1667}
\textsc{D.~Schwenter.}
\textit{Geometriae Practicae novae et auctae Libri IV \ldots
mit vielen nutzlichen Additionen und neuen Figuren vermehret durch
G.~A.~Böcklern}. Nürnberg.

{\footnotesize
``Von Austheilung der Figuren in gleiche und ungleiche Theil,'' pp.~269--279;
p.~350; the problem on this last page is taken from \textsc{B.~Bramer}, \textit{Trigonometria
planarum mechanica}, Marpurg, 1617, p.~99. ``Von Austheilungen der Aecker
Wiesen,\ldots '' pp.~567--583.}

% -----File: 093.png---Folio 81------------------------------------------------------

\AppItem{1674}{1674}
\textsc{C.~F.~M.~Deschales.}
\textit{Cursus seu mundus mathematicus}. Lugduni.

{\footnotesize
``De figurarum planarum divisione,'' \textsc{i},~371--381; second edition, 1690.}

\AppItem{1676}{1676}
\textsc{I.~Newton.}
\textit{Arithmetica Universalis}. Cantabrigiae, \textsc{MDCCVII}.

{\footnotesize
Prob.\ \textsc{X}, p.~126 (Prob.\ \textsc{XX}, pp.~254--255 of the 1769 edition). This problem was
discussed in a lecture delivered October, 1676 (see \textit{Correspondence of Sir Isaac Newton
and Professor Cotes}\ldots by J.~Edleston, London, 1850, p.~xciii).}

\AppItem{1684}{1684}
\textsc{T.~Strode.}
\textit{A Discourse of Combinations, Alternations and Aliquod
Parts} by John Wallis. London, 1685.

{\footnotesize
On pages 163--164 is printed a letter, dated Nov.~1684, from Strode to Wallis. It
discusses two problems on divisions of a triangle.}

\AppItem{1687}{1687}
\textsc{J.~Bernoulli.} ``Solutio algebraica problematis de
quadrisectione trianguli scaleni, per duas normales rectas.''
\textit{Acta Eruditorum}, 1687, pp.~617--623.

\footnotesize
Also in \textit{Opera}, Genevae, 1744, I,~328--335; see further II,~671. In the solution of this
question Bernoulli is led to the intersection of a conic and a curve of the fourth degree,
that is, to an equation of the eighth degree. And yet, in the seventh edition of
\textsc{Rouché} et \textsc{Comberousse},
\textit{Traité de Géométrie}, Paris, 1900, we find Problem 453 is:
``Partager un triangle quelconque en quatre parties équivalentes par deux droites
perpendiculaires entre elles!'' The problem was solved by L'Hospital before 1704,
the year of his death, in a posthumous work, \textit{Traité analytique des Sections coniques},
Paris, 1707, pp.~400--407. As the result of correspondence in \textit{L'Intermédiaire des
Math}., tomes \textsc{i--vii}, 1894--1900, Questions 3 and 587, Loria wrote the history of the
problem: ``Osservazioni sopra la storia di un problema pseudo-elementare.'' \textit{Bibl.\
Math}., 1903 (3), \textsc{iv}, 48--51. Leibnitz's name appears in this connection. See note on
1645---C.~Huygens.
\small

\AppItem{1688}{1688}
\textsc{J.~Ozanam.}
\textit{L'usage du compas de proportion expliqué et demontré d'une
manière \DPtypo{court}{courte}
et facile, et augmenté d'un Traité de la division des champs}.
Paris.

{\footnotesize
``Division des champs,'' pp.~89--138. \DPtypo{Edition}{Ouvrage}   %% see books.google.com
revu, corrigé et entièrement refondu
par J.~G.~Garnier. Paris, 1794, pp.~165--257.}

\AppItem{1694}{1694}
\textsc{S.~Le Clerc.} \textit{Traité de Géométrie sur le terrain} at end of
\textit{Géométrie pratique, ou pratique de la géométrie sur le papier et sur le terrain}.
Amsterdam.

\AppItem{1699}{1699}
\textsc{J.~Ozanam.}
\textit{Cours de mathématique}, nouv.\ éd. tome 3. Paris.

{\footnotesize
Pages 23--64. German translation: \textit{Anweisung, die geradlinichten Figuren nach
einen gegebenen Verhältniss ohne Rechnung zu theilen}. Frankfurt u. Leipzig, 1776.}

\AppItem{1704}{1704}
\textsc{Guisnée.}
\textit{Application de l'algèbre à la géométrie}. Paris.
\nopagebreak

{\footnotesize
Although the ``approbation'' signed by Fontenelle is dated ``15 Juillet 1704'' the
work was first published in 1710; second edition ``revûe, corrigée et considérablement
augmentée par l'auteur,'' Paris, 1733, pp.~42--47; analytic discussion only.}

\AppItem{1739}{1739}
\textit{l'abbé}~\textsc{Deidier.}
\textit{La science des géométres} (\textit{sic})
\textit{ou la théorie
et la pratique de la géométrie}. Paris.

{\footnotesize
``De la géodésie ou division des champs,'' pp.~279--320; divisions of triangles, rectangles,
trapeziums, polygons.}

\AppItem{1740}{1740}
\textsc{N.~Saunderson.}
\textit{Elements of Algebra in ten books}, vol.~2. Cambridge.

{\footnotesize Pages 546--554.}

\AppItem{1747}{1747}
\textsc{T.~Simpson.} \textit{Elements of Plane Geometry}. London.

{\footnotesize Pages~151--152; new ed., London, 1821, pp.~207--208; taken from Newton (1676).}

% -----File: 094.png---Folio 82------------------------------------------------------
\AppItem{1748}{1748}
\textsc{L.~Euler.}
\textit{Introductio in analysin infinitorum}. Tomus secundus. Lausanne.

{\footnotesize
Chapter 22: ``Solutio nonnullorum problematum ad circulum pertinentium.''
Three of the eight problems which Euler here discusses by the method of trial and
error, and tables of circular arcs and logarithmic sines and tangents, are of interest to
us. These are: Problem 2, ``To find the sector of the circle $\mathit{ACB}$ which is divided
by the chord $\mathit{AB}$ into two equal parts, so that the triangle $\mathit{ACB}$ shall be equal to the
segment $\mathit{AEB}$.'' Problem 4, ``Given the semi-circle $\mathit{AEDB}$, to draw from the point
$A$ a chord $\mathit{AD}$ which will divide the semi-circle into two equal parts.'' Problem 5,
``From a point $A$ of the circumference of a circle, to draw two chords $\mathit{AB}$, $\mathit{AC}$ which
shall divide the area of the circle into three equal parts.'' (Heron, \textit{cf.}\ Art.~50.)
Gregory (1840) considers these problems at the close (pp.~186--188) of his Appendix.

For other editions of Euler's ``Introductio,'' tomus 2, see
\textit{Verzeichnis der Schrif\-ten Leonhard Eulers.}
Bearbeitet von G.~Eneström. Erste Lieferung, Leipzig,~1910.}


\AppItem{1752}{1752}
\textsc{T.~Simpson.}
\textit{Select Exercises for young proficients in the mathematics.} London.

{\footnotesize
Problem \textsc{xlii}, pp.~145--6; new ed.\ by J.~H.~Hearding.  London, 1810, pp.~148--9.}

\AppItem{1754}{1754}
\textsc{J.~le~R.~D'Alembert.}
\textit{Encyclopédie ou Dictionnaire raisonné des
sciences\ldots mis en ordre et publié par M.~Diderot\ldots; et quant à la partie
mathématique par M.~d'Alembert.} Paris.

{\footnotesize
Article ``Géodésie'';
mostly descriptive of methods of Guisnée (1704) and Clerc (1694).}

\AppItem{1768}{1768}
\textsc{J.~A.~Euler.} ``Auflösung einiger geometrischen Aufgaben,''
\textit{Abhandlungen der Churfürstlich-baierischen Akademie der Wis\-sen\-schaf\-ten},
v, 165--196.

{\footnotesize
Erste Aufgabe, pp.~167--182: ``Man soll zeigen, wie eine jede geradlinichte Figur
durch Parallellinien in eine gegebene Anzahl gleicher Theile zerschnitten werden
kann.'' Zweite Aufgabe, pp.~182--187: ``Eine Zirkel-fläche durch parallellinien in
eine gegebene Anzahl gleicher Theile zu zerschneiden.'' Dritte Aufgabe, pp.~187--196:
``Die Höhe und Grundlinie einer aufrecht-stehenden geschlossenen Parabelfläche ist
gegeben, man soll dieselbe durch Parallellinien in $n$ gleiche Theile zerschneiden.''
Discussion mostly analytic.}

\AppItem{1772(?)}{1772}
\textit{J.~H.~Lamberts deutscher gelehrter Briefwechsel.}
Herausgegeben von Joh.\ Bernoulli. Band 2, Berlin, 1782.

{\footnotesize
Pages 412--13, undated fragment of a letter from Lambert to J.~E.~Silberschlag.
Analytic solution by quadratic equation of the problem: ``Ein Feld $\mathit{ABCD}$ welches
in $\mathit{ABFE}$ Wiesen, in $\mathit{EFCD}$ Ackerfeld ist,
soll durch eine gerade Linie $\mathit{KM}$ so
getheilt werden, dass so wohl die Wiesen als das Ackerfeld in beliebiger Verhältniss
getheilt werde.''
[$\mathit{ABCD}$ is a quadrilateral and $\mathit{EF}$ is a straight line segment joining
points on the opposite sides $\mathit{AD}$, $\mathit{BC}$ respectively.]

In the \textit{Journal of the Indian Mathematical Society}, 1914,
\textsc{vi}, 159, N.~P. Pan\-dya
proposed as Question 563: ``Given two quadrilaterals in the plane of the paper show
how to draw a straight line bisecting them both.'' A solution by means of common
tangents to hyperbolas was offered in 1915, \textsc{vii}, 176.}

\AppItem{1783}{1783}
\textsc{J.~T.~Mayer.}
\textit{Gründlicher und ausführlicher Unterricht zur praktischen Geometrie,}
3.~Teil. Göttingen.

{\footnotesize
Pages 215--303: ``Theilung der Felder durch Rechnung, Theilung der Felder durch
blose Zeichnung, Anwendung der Theilungsmethoden auf mancherley, in gemeinen
Leben vorkommende Fälle''; dritte Auflage, 1804, pp.~232--337.}

\AppItem{1793}{1793}
\textsc{J.~W.~Christiani.}
\textit{Die Lehre von der geometrischen und ökonomischen
Vertheilung der Felder, nach der Dänischen Schrift des N.~Morville
bearbeitet von J.~W.~Christiani.} Preface by A.~G.~Kästner. Göttingen.

\AppItem{1795}{1795}
\textit{Gentleman's Diary}, London.

{\footnotesize
No.~54, 1794, p.~47, Question 691 by J.~Rodham: ``Within a given triangle to find
% -----File: 095.png---Folio 83------------------------------------------------------
a point thus, that if lines be drawn from it to cut each side at right angles, the three
parts into which the triangle thus becomes divided, shall obtain a given ratio.'' Solution
by hyperbolas in No.~55, 1795, pp.~37--38.
See also Davis's edition of the \textit{Gentleman's Diary}, vol.~3, London, 1814, pp.~233--4.}

\AppItem{1801}{1801}
\textsc{L.~Puissant.}
\textit{Recueil de \DPtypo{divers}{diverses}
propositions de géométrie résolues ou
démontrées par l'analyse algébrique suivant les principes de Monge et de
Lacroix}. Paris.

{\footnotesize
Pages 33--36; German ed., Berlin, 1806; second French ed., Paris, 1809,
pp.~107--111; third ed., Paris, 1824, pp.~139--142.}

\AppItem{1805}{1805}
\textsc{M.~Hirsch.}
\textit{Sammlung geometrischer Aufgaben}, Erster Theil. Berlin.

{\footnotesize
``Theilung der Figuren durch Zeichnung,'' pp.~14--25; ``Theilung der Figuren
durch Rechnung,'' pp.~42--53; Reprint, 1855; English edition translated by J.~A.~Ross
and edited by J.~M.~F.~Wright. London, 1827.}

\AppItem{1807}{1807a}
\textsc{A.~Bratt.}
\textit{Problema geometricum triangulum datum a dato puncto in
2 partes aequales secandi}. Greifswald.

{\footnotesize
This title is taken from \textsc{C.~G.~Kayser},
\textit{Bücher-Lexicon}, Erster Teil, Leipzig, 1834.}

\AppItem{1807}{1807b}
\textsc{J.~P.~Carlmark.}
\textit{Triangulus datus a dato puncto in 2 partes aequales secandus}. Greifswald.

{\footnotesize
This title and the next two are taken from \textsc{E.~Wölffing},
\textit{Math.\ Bücherschatz}, 1903.}

\AppItem{1809}{1809}
\textsc{J.~Kullberg.}
\textit{Problema geometricum triangulum datum e quovis dato
puncto in 2 partes aequales secandi}. Diss.\ Lund.

\AppItem{1810}{1810}
\textsc{J.~Kullberg.}
\textit{Problema geometricum triangulum quodcunque datum in
2 aequales divisum iterum in partes aequales ita secandi, ut rectae secantes
angulum constituant rectum}. Diss.\ Upsala.

\AppItem{1811}{1811}
\textsc{J.~P.~Grüson.}
\textit{Geodäsie oder vollständige Anleitung zur geometrischen
und ökonomischen Feldertheilung}. Halle.

\AppItem{1819}{1819}
\textsc{L.~Bleibtreu.}
\textit{Theilungslehre oder ausführliche Anleitung},
\textit{jede Grundfläche auf die zweckmässigste Art} \ldots
\textit{geometrisch zu theilen}. Frankfurt am Main.

\AppItem{1821}{1821}
\textsc{J.~Leslie.}
\textit{Geometrical Analysis and Geometry of Curve Lines}
Edinburgh.

{\footnotesize Pages 64--66.}

\AppItem{1823}{1823}
\textsc{A.~K.~P.~von Forstner.}
\textit{Sammlung systematisch geordneter und
synthetisch \DPtypo{aufgelöseter}{aufgelöster}
geometrischer Aufgaben}. Berlin.

{\footnotesize ``Theilung der Flächen, mittelst der Proportion und der Aehnlichkeit,''
pp.~310--371.}

\AppItem{1827}{1827}
\textit{Correspondance mathématique et physique}
publié par A.~Quetelet, tome \textsc{iii}.

{\footnotesize
Page 180: ``On donne dans un plan un angle et un point,
et l'on demande de faire
passer par le point une droite qui coupe les cotés de l'angle,
de manière que l'aire
interceptée soit de grandeur donnée.'' Solution by Verhulst,
pp.~269--270. Answer
by Bobillier, tome \textsc{iv}, pp.~2--3.
Generalizing his solution, he gets the result: ``tous
les plans tangens d'un hyperboloïde à deux nappes,
interceptent sur le cône asymptotique
des volumes équivalens.'' Compare note 117.}

\AppItem{1831}{1831a}
\textsc{P.~L.~M.~Bourdon.}
\textit{Application de l'algèbre à la géométrie comprenant
la géométrie analytique à deux et à trois dimensions}, troisième édition.
Paris.

{\footnotesize
Pages 46--54; 5$^e$~éd., Paris, 1854, pp.~33--41; 8$^e$~éd.\ rev.\ par
Darboux, Paris, 1875, pp.~30--38. Analytic discussion only.}
% -----File: 096.png---Folio 84------------------------------------------------------

\AppItem{1831}{1831b}
\textsc{H.~v.~Holleben}, und \textsc{P.~Gerwien.}
\textit{Geometrische Analysis}. Berlin, 2~Bde, 1831--1832.

{\footnotesize ``Theilungen,'' \textsc{i},~184--191; \textsc{ii},~144--151.}

\AppItem{1837}{1837}
\textsc{G.~Ritt.}
\textit{Problèmes d'applications de l'algèbre à la
géométrie avec les solutions développées}, 2$^e$~partie. Paris.

{\footnotesize Pages 108--109.}

\AppItem{1840}{1840}
\textsc{O.~Gregory.}
\textit{Hints theoretical, elucidatory and practical, for the use
of Teachers of elementary Mathematics and of self-taught students; with
especial reference to the first volume of Hutton's course and Simson's Euclid,
as Text-Books. Also a selection of miscellaneous tables, and an Appendix
on the geometrical division of plane surfaces}. London.

{\footnotesize ``Appendix: Problems relative to the division of Fields and other surfaces,''
pp.~158--188; partly taken from Hirsch~(1805). See also Euler~(1748).}

\AppItem{1844}{1844}
\textsc{Dreser.} \textit{Die Teilung der Figuren}. Darmstadt.

{\footnotesize This title is taken from \textsc{E.~Wölffing}, \textit{Math.\ Bücherschatz}, 1903.}

\AppItem{1847}{1847}
\textsc{R.~Potts.} \textit{An appendix to the larger edition of
Euclid's Elements of Geometry; containing\ldots Hints for the solution
of the Problems}\ldots Cam\-bridge and London.\DPnote{Would not hyphenate Cambridge without the hint. ??}

{\footnotesize Ex.~91, pp.~72--73.}

\AppItem{1852}{1852a}
\textsc{H.~Ch.~de~La Frémoire.}
\textit{Théorèmes et Problèmes de Géométrie
élémentaire}, \DPtypo{second}{seconde}
éd.\ revue et corrigée par E.~Catalan. Paris.

{\footnotesize
Pages 107--108; 6$^e$ éd.\ par Catalan, Paris, 1879, pp.~190--191.}

\AppItem{1852}{1852b}
\textsc{F.~Rummer.}
\textit{Die Verwandlung und Theilung der Flächen in einer
Reihe von Constructions- u.\ Berechnungs-Aufgaben}. Mit 3 Steintafeln.
Heidelberg. 6~+~90 pp.

\AppItem{1855}{1855}
\textsc{P.~Kelland.} ``On Superposition.''
\textit{Transactions of the Royal Society of Edinburgh},
1885, \textsc{xxi}, 271--273 + 1 pl.

{\footnotesize
This paper deals, for the most part, with solutions of the following problem proposed
to Professor Kelland by Sir John Robison: ``From a given square one quarter is cut
off, to divide the remaining gnomon into four such parts that they shall be capable of
forming a square.'' In the \textit{Transactions}, 1891,
\textsc{xxxvi}, 91--95, + 2 pls.,\ Robert Brodie
has a paper entitled ``Professor Kelland's Problem on Superposition.''}

\AppItem{1857}{1857}
\textsc{E.~Catalan.}
\textit{Manuel des Candidats \DPtypo{a}{à}
l'école polytechnique}. Paris,
Tome I.

\footnotesize
Pages 233--4: ``To divide a circle into two equal parts by means of an arc with its
centre, \textit{A}, on the circumference of the given circles.''
This is stated by \textsc{A.~Rebière}
(\textit{Mathématique et Mathématiciens},
2$^e$~éd.,\ Paris, 1893, p.~519) under the form:
``Quelle doit être la longueur de la longe d'un cheval pour qu'en
la fixant au contour d'un pré circulaire l'animal ne puisse
tondre que la moitié du pré?''

The solution of this problem leads to a transcendental equation

\begin{equation*}
\sin x - x \cos x = \frac{\pi}{2},
\end{equation*}

where $x$ is the angle under which the points of section of the circumferences are seen
from \textit{A}. Catalan finds $x = 109^\circ 11' 18''$, correct to within a second of arc.

Cf.\ \textit{L'Intermédiaire des Mathématiciens}, 1914,
Question 4327, \textsc{xxi}, 5, 69, 90, 115, 180.
\small

\AppItem{1863}{1863}
\textsc{J.~ McDowell.}
\textit{Exercises on Euclid and in Modern Geometry}.
Cambridge.

{\footnotesize No.~157, pp.~145--6; 3rd ed.\ 1881, p.~118.}

% -----File: 097.png---Folio 85------------------------------------------------------
\AppItem{1864}{1864a}
\textit{Educational Times Reprint},
Vols.\ 1, 40, 44, 66, 68, 69; new series, Vol.\ 1; 1864--1910.

{\footnotesize
The problems here solved are Euclid's 19, 20, 26, 27: No.~1457 (\textsc{i}, 49, old edition,
1864) proposed by R.~Palmer; solution by Rutherford who states that it was also
published in Thomas Bradley's \textit{Elements of Geometrical Drawing}, 1861---Nos.~7336 and
7369 (\textsc{xl}, 39, 1884) proposed by W.~H.~Blythe and A.~H.~Curtis; solutions by
G.~Heppel and Matz---No.~8272 (\textsc{xliv}, 92, 1886) proposed by E.~Perrin; solution by
D.~Biddle---No.~12973 (\textsc{lxvi}, 29, 1897) proposed by Radhakrishnan; solution by
I.~Arnold---No.~13460 (\textsc{lxviii}, 35, 1898) proposed by I.~Arnold; solution by
W.~S.~Cooney, etc.---No.~13647 (\textsc{lxix}, 42, 1898) proposed by I.~Arnold; solution by
W.~C.~Stanham---No.~16747 (new series \textsc{xviii}, 46, 1910) proposed by I.~Arnold;
solution by proposer, by Euclid's \textit{Elements} Bk~\textsc{i}.}

\AppItem{1864}{1864b}
\textsc{H.~Hölscher.}
\textit{Anleitung zur Berechnung und Teilung der Polygone
bei rechtwinkligen Koordinaten}. Berlin and Charlottenburg.

{\footnotesize
This work and the two following are representative of those which treat of
Divisions of Figures by computation, rather than by graphical methods:
(1) \textsc{F.~G.~Gauss},
\textit{Die Teilung der Grundstücke, insbesondere unter Zugrundelegung Koordinaten},
2 Auflage, Berlin, 1890;
(2) \textsc{L.~Zimmerman}, \textit{Tafeln für die Teilung der Dreiecke,
Vierecke, und Polygone}, Zweite vermehrte und verbesserte Auflage, Liebenwerda,
1896; 118+64 pp.}

\AppItem{1870}{1870}
\textsc{F.~Lindman.} ``Problema geometricum.''
\textit{Archiv der Math.\ u.\ Phys}.
(Grunert), Bd~51, 1870, pp.~247--252.

\AppItem{1879}{1879}
\textsc{P.~M.~H.~Laurent.}
\textit{Traité d'algèbre à l'usage des candidats\ldots.}
Troi\-si\-ème édition. Paris.

{\footnotesize
Tome I, p.~191:
``To divide a triangle into two equal parts by the shortest possible line.''
Solutions in \textit{L'Intermédiaire des Mathématiciens}, 1902, IX, 194--5. See also
F.~G.~M., \textit{Exercices de Géométrie}, Cinquième édition. Tours et Paris, 1912, p. 802.}

\AppItem{1892}{1892}
\textsc{H.~S.~Hall} and \textsc{F.~H.~Stevens.}
\textit{Key to the Exercises and Examples contained in a Text-Book of Euclid's Elements}.
London.

{\footnotesize Ex.\ 7, 8, 10, 11, pp.~163--164.}

\AppItem{1894}{1894}
\textsc{G.~E.~Crawford.} ``Geometrical Problem.''
\textit{Proc.\ Edinb.\ Math.\ Soc.}, Vol.~13, 1895, p.~36.

{\footnotesize Paper read Dec.~14, 1894.}

\AppItem{1899}{1899}
\textsc{W.~J.~Dilworth.} \textit{A New Sequel to Euclid}. London.

{\footnotesize Ex. \textsc{xxxv}, p.~190.}

\AppItem{1901}{1901}
\textsc{A.~Larmor.}
\textit{Geometrical Exercises from Nixon's `Euclid Revised.'}
Oxford.

{\footnotesize Ex.~15, p.~122.}

\AppItem{1902}{1902}
\textsc{C.~Smith.}
\textit{Solution of the Problems and Theorems in Smith and
Bryant's Elements of Geometry}. London.

{\footnotesize Ex.~121, pp.~177--178; T.~Simpson's solution and another.}

\AppItem{1910}{1910a}
\textsc{H.~Flükiger.}
\textit{Die Flächenteilung des Dreiecks mit Hilfe der Hyperbel.}
Diss.\ Bern. 50 pp. + 3 plates.

\AppItem{1910}{1910b}
\textsc{R.~Zdenek.} ``Halbierung der Dreiecksfläche.'' Wien,
\textit{Zeitschrift für das Realwesen}, Jahrgang \textsc{xxxv}, Heft~10, 8~pp.

{\footnotesize Discussion by projective geometry leading to hyperbolic arcs.}

\AppItem{1911}{1911}
\textsc{D.~Biddle.} Problem 17197,
\textit{Educational Times}, London, November,
\textsc{lxiv}, 475.

{\footnotesize
``Divide a square into five right-angled triangles, the areas of which shall be in
arithmetic progression.'' Solutions in the Educational Times Reprint, new series,
\textsc{xxvi, iii}, 1914.}

\end{Description}

% -----File: 098.png---Folio 86------------------------------------------------------
\backmatter
\label{indexpage}
\phantomsection\pdfbookmark[0]{Index of Names}
                              {Index of Names}



\fancyhead[C]{\Heading{INDEX OF NAMES}}
\thispagestyle{empty}
\renewcommand{\indexname}{\large{\centerline{INDEX OF NAMES}}}
\renewcommand{\indexhook}{\textit{In the following list, references are
 given to paragraph
and footnote} (\textit{=~n.})\ \textit{numbers, except in the case of
the Appendix} (\textit{=~App.})\ \textit{where the numbers
are the years in the chronological list. \; App.,\ without a number,
 refers to
the introductory paragraphs on page \pageref{appendix}.}\bigskip}
\begin{theindex}
\footnotesize

\item Abraham bar Chijja ha Nasi or Ab\-ra\-ham Sa\-va\-sor\-da \ParRef{18}, \FnRef{53}
\item Abû Ishâq b.\ `Abdallâh \ParRef{19}
\item Abû Jûsuf Ja`qûb b.\ Ishâq b.\ el-Sabbah el-Kindî, \textit{see} el-Kindî
\item Abû'l Wefâ \ParRef{19}, \FnRef{78}
\item Abû Muh.\ b.\ Abdalbâqî el Ba\.{g}dâdî el-Faradî, \textit{see} el-Ba\.{g}dâdî
\item Abû Muh.\ el-Hasan b.\ `Obeidallâh b.\ Soleimân, \textit{see} el-Hasan
\item Ainsworth, R. \FnRef{28}
\item Albategnius = Al-Battânî = el-Bat\-tânî, \textit{see there}
\item Al-Kindî = el-Kindî
\item Anderson, A. \AppRef{1619}{\DPtypo{1616}{1619}}
\item Antiphon \FnRef{116}
\item Apollonius of Perga \ParRef{19}, \ParRef{21}, \FnRef{103}, \FnRef{111}
\item Archimedes \ParRef{19}, \FnRef{601}, \FnRef{83}, \FnRef{101},
      \FnRef{103}, \FnRef{109}, \FnRef{111}
\item Aristarchus of Samos \FnRef{111}
\item Aristotle \FnRef{116}
\item Armagh, Archbishop of, \textit{see} Ussher, J.
\item Arnold, I. \AppRef{1864a}{1864}
\item Ashmole \FnRef{22}
\item Athelhard of Bath \FnRef{15}
\item Ayton, W.~A. \FnRef{14}
\indexspace
\item Bachmann, P. \AppRef{appendix}{}
\item Benedetti, G.~B. \AppRef{1585}{1585}
\item Bernoulli, J.    \AppRef{1645}{1645},\AppRef{1687}{1687},\AppRef{1772}{1772(?)}
\item Biddle, D.       \AppRef{1864a}{1864},\AppRef{1911}{1911}
\item Bleibtreu, L.    \AppRef{1819}{1819}
\item Blythe, W.~H.    \AppRef{1864a}{1864}
\item Bobillier, E.~E. \FnRef{117}, \AppRef{1827}{1827}
\item Böcklern, G.~A.  \AppRef{1667}{1667}
\item Boncompagni, B.  \ParRef{10}, \ParRef{13},
      \FnRef{40},  \FnRef{119},  \AppRef{appendix}{}, \AppRef{1560}{1560}
\item Bossut, C.       \AppRef{1609}{1609}
\item Bourdon, P.~L.~M. \AppRef{1831a}{1831}
\item Bradley, T.      \AppRef{1864a}{1864}
\item Bramer, B.       \AppRef{1667}{1667}
\item Bratt, A.        \AppRef{1807a}{1807}
\item Breton (de Champ), P. \FnRef{88}
\item Brodie, R.       \AppRef{1855}{1855}
\item Bryant, S.       \AppRef{1902}{1902}
\indexspace
\item Campanus, J. \ParRef{2}, \ParRef{5}, \FnRef{11}, \FnRef{15}
\item Candale, \textit{see} Flussates
\item Cantor, M. \ParRef{2}, \FnRef{8}, \FnRef{39}, \FnRef{53},
      \FnRef{55}, \FnRef{62}, \FnRef{63}, \FnRef{82},
      \FnRef{103}, \FnRef{107}, \FnRef{120}, \AppRef{1547b}{1547}, \AppRef{1609}{1609}
\item \DPtypo{Cardan, G.}{Cardano, M.~H.} \AppRef{1547b}{1547}
\item Cardinael, S. \AppRef{1612}{1612}
\item Carlmark, J.~P. \AppRef{1807b}{1807}
\item Casley, D. \ParRef{4}, \FnRef{20}
\item Cassini, J. \AppRef{1609}{1609}
\item Catalan, E. \AppRef{1852a}{1852}, \AppRef{1857}{1857}
\item Chasles, M.  19, \FnRef{65}, \FnRef{85}, \FnRef{88},
      \FnRef{111}
\item Christiani, J.~W. \FnRef{67}, \AppRef{1793}{1793}
\item Clairaut, A.~C. \AppRef{1609}{1609}
\item Clavius, C. \AppRef{1604}{1604}
\item Clerc, S. le \AppRef{1694}{1694}, \AppRef{1754}{1754}
\item Comberousse, C. de \AppRef{1687}{1687}
\item Commandinus, F. \ParRef{2}, \ParRef{5}, \ParRef{1}5, \FnRef{10}, \FnRef{11},
      \FnRef{35}, \FnRef{49}, \FnRef{50}, \FnRef{88}
\item Copernicus 19, \FnRef{35}, \FnRef{111}
\item Cossali, P. \FnRef{119},  \AppRef{appendix}{}, \AppRef{1560}{1560}
\item Cotes, R. \AppRef{1676}{1676}
\item Cotton, R.~B. \ParRef{4}, \FnRef{21}
\item Cottonian MSS.  \ParRef{2}, \ParRef{3}, \ParRef{4}, \ParRef{5},
      \FnRef{18}, \FnRef{19},
      \FnRef{20}, \FnRef{21}
\item Cowley, A. \ParRef{6}
\item Cratfield, W. \ParRef{4}
\item Crawford, G.~E. \AppRef{1894}{1894}
\item Crelle \FnRef{117}
\item Cresswell, D. \FnRef{110}
\item Curtis, A.~H. \AppRef{1864a}{1864}
\item Curtius, S. \AppRef{1612}{1612}
\item Curtze, M. \ParRef{18}, \FnRef{54}, \FnRef{55}
\indexspace
\item D'Alembert, J. le R. \AppRef{1754}{1754}
\item Darboux, G. \AppRef{1831a}{1831}
\item Davis, A. \AppRef{1795}{1795}
\item Dee, J. \ParRef{2}, \ParRef{3}, \ParRef{4}, \ParRef{5}, \ParRef{6}, \ParRef{8},
      \ParRef{9}, \ParRef{18}, \ParRef{19}, \FnRef{10}, \FnRef{11},
      \textit{n}.~14, \FnRef{24}, \FnRef{25},
      \FnRef{29}, \FnRef{35}
\item Deidier, \textit{l'abbé} \AppRef{1739}{1739}
\item Deschales, C.~F.~M. \AppRef{1674}{1674}
\item Diderot, D. \AppRef{1754}{1754}
\item Diels, H. \FnRef{116}
\item Diesterweg, W.~A. \FnRef{111}
\item Dilworth, W.~J. \AppRef{1899}{1899}
\item Dou, I. \AppRef{1612}{1612}
\item Dreser \AppRef{1844}{1844}
\indexspace
\item \DPtypo{Edelston}{Edleston}, J. \AppRef{1676}{1676}
\item el-Ba\.{g}dâdî\footnote{
All the ``Muhammeds'' of Bagdad referred to in this volume are here supposed to be
indicated by this single name.}  % end footnote
      \ParRef{2}, \ParRef{3}, \ParRef{4}, \ParRef{5}, \ParRef{6},
       \ParRef{13}, \ParRef{14}, \ParRef{15}, \ParRef{19},
       \FnRef{11},
% -----File: 099.png---Folio 87------------------------------------------------------
      \FnRef{12}, \FnRef{15}, \FnRef{17}, \FnRef{21},
      \FnRef{29}, \FnRef{33}, \FnRef{35}, \AppRef{appendix}{}
\item el-Battânî  \ParRef{19}, \FnRef{35}
\item el-Bûz\v{g}ânî, \textit{see} Abû'l Wefâ
\item el-Hasan  \ParRef{19}
\item el-Kindî  \ParRef{19}, \FnRef{35}
\item Eneström,~G. \FnRef{55}, \AppRef{1748}{1748}
\item Euclid  \ParRef{1}, \ParRef{2}, \ParRef{5}, \ParRef{6}, \ParRef{7}, \ParRef{8},
      \ParRef{9}, \ParRef{10}, \ParRef{11}, \ParRef{12}, \ParRef{13}, \ParRef{16},
      \ParRef{17}, \ParRef{18}, \ParRef{19}, \ParRef{20}, \ParRef{21}, \ParRef{34},
      \ParRef{50}, \ParRef{57},
   \FnRef{1},   \FnRef{3},   \FnRef{5},   \FnRef{6},   \FnRef{9},  \FnRef{11},
   \FnRef{15},  \FnRef{30},  \FnRef{34},  \FnRef{35},  \FnRef{36}, \FnRef{38},
   \FnRef{45},  \FnRef{47},  \FnRef{50},  \FnRef{56},  \FnRef{57}, \FnRef{81},
   \FnRef{83},  \FnRef{85},  \FnRef{87},  \FnRef{88},  \FnRef{91}, \FnRef{94},
   \FnRef{95},  \FnRef{98},  \FnRef{103}, \FnRef{105}, \FnRef{106},
   \FnRef{108}, \FnRef{109}, \FnRef{111}, \FnRef{113}, \FnRef{116},
   \FnRef{1181}, \AppRef{appendix}{}, \AppRef{1560}{1560}, \AppRef{1840}{1840}, \AppRef{1864a}{1864}
\item Eudemus \FnRef{116}
\item Euler, J.~A. \AppRef{1768}{1768}
\item Euler, L. \AppRef{appendix}{}, \AppRef{1748}{1748}, \AppRef{1840}{1840}
\item Eutocius \FnRef{111}
\indexspace
\item Fabricius, J.A.  \ParRef{19}, \FnRef{68}
\item Favaro, E.~A.  \ParRef{2}, \ParRef{6}, \ParRef{10}, \ParRef{13}, \ParRef{15},
      \ParRef{19}, \ParRef{29},
   \FnRef{6}, \FnRef{32}, \FnRef{42}, \FnRef{52}, \FnRef{73}, \FnRef{97}, \FnRef{100}
\item Ferrari \AppRef{1547a}{1547}
\item F.~G.~M. \AppRef{1879}{1879}
\item Fibonaci, \textit{see} Pisano Leonardo
\item Flükiger, H. \AppRef{1910a}{1910}
\item Flussates or Foix, \textit{i.~e.} François de Foix-Candalle \FnRef{11}
\item Fontenelle \AppRef{1704}{1704}
\item Forstner, A.~K.~P. von \AppRef{1823}{1823}
\item Frankland, W.~B. \FnRef{103}
\item Friedlein, G. \FnRef{1}
\item Frisch, C. \AppRef{1609}{1609}
\indexspace
\item Gardiner, M. \FnRef{111}
\item Garnier, J.~G. \AppRef{1688}{1688}
\item Gauss, C.~F. \FnRef{118}
\item Gauss, F.~G. \AppRef{1864b}{1864}
\item Gerwien, P. \AppRef{1831b}{1831}
\item Gherard of Cremona  \ParRef{4}, \ParRef{11}, \ParRef{19}
\item Giordani, E. \AppRef{1547b}{1547}
\item Grabow, M.~G. \FnRef{111}
\item Gregory, D. 6, \FnRef{11}, \AppRef{1609}{1609}
\item Gregory, O. \AppRef{1748}{1748}, \AppRef{1840}{1840}
\item Grüson, J.~P. \AppRef{1811}{1811}
\item Grunert, J.~A.  \ParRef{19}, \FnRef{72}, \AppRef{1870}{1870}
\item Gudermann, C. \FnRef{117}
\item Guisnée \AppRef{1704}{1704}, \AppRef{1754}{1754}
\item Gutman, J. \AppRef{1574a}{1574}
\indexspace
\item Hall, H.~S. \AppRef{1892}{1892}
\item Halley, E. \FnRef{111}
\item Halliwell, J.O. 4, \FnRef{26}
\item Hankel, H.  \ParRef{2}, \ParRef{19}, \FnRef{4}, \FnRef{71}, \FnRef{111}
\item Harleian MSS.  \ParRef{6}, \FnRef{21}, \FnRef{22}, \FnRef{31}
\item Hearding, J.~H. \AppRef{1752}{1752}
\item Heath, T.L.  \ParRef{2}, \ParRef{3}, \ParRef{6},
      \FnRef{3}, \FnRef{15}, \FnRef{601},
      \FnRef{81}, \FnRef{83}, \FnRef{87}, \FnRef{88}, \FnRef{99},
      \FnRef{101}, \FnRef{103}, \FnRef{108}, \FnRef{111},
      \FnRef{112}, \FnRef{116}
\item Heiberg, J.~L.  \ParRef{2}, \ParRef{6}, \ParRef{7}, \ParRef{8},
      \FnRef{5}, \FnRef{32}, \FnRef{36},
      \FnRef{601}, \FnRef{83}, \FnRef{111}
\item Heilbronner, J.~C.  \ParRef{6}, \ParRef{19}, \FnRef{33}, \FnRef{69}
\item Heppel, G. \AppRef{1864a}{1864}
\item Herigone, P. \FnRef{111}
\item Hermann, J. \AppRef{1609}{1609}
\item Heron of Alexandria  \ParRef{18}, \ParRef{20}, \ParRef{21},
      \FnRef{81}, \FnRef{82}, \FnRef{83}, \FnRef{84}
\item Hippocrates \FnRef{116}
\item Hirsch, M. \AppRef{1805}{1805}
\item Hölscher, H. \AppRef{1864b}{1864}
\item Holleben, H. von \AppRef{1831b}{1831}
\item Hooper, S. \FnRef{20}
\item Hultsch, F.  \ParRef{2}, \FnRef{9}, \FnRef{36}, \FnRef{85}, \FnRef{109}
\item Hutton, C.  \AppRef{appendix}{}, \AppRef{1840}{1840}
\item Huygens, C.  \AppRef{appendix}{}, \AppRef{1612}{1612}, \linebreak[2]
      \AppRef{1645}{1645}, \AppRef{1657}{1657}, \AppRef{1687}{1687}
\indexspace
\item Ishâq b. Hunein b. Ishâq el-`Ibâdî, Abû Ja`qûb  \ParRef{19}
\indexspace
\item Jacobi, C.~J.~A. \FnRef{111}
\item Jordanus Nemorarius  \ParRef{18}, \FnRef{58}
\indexspace
\item Kästner, A.~G.  \ParRef{2}, \ParRef{3}, \ParRef{19}, \FnRef{16}, \FnRef{19},
      \FnRef{67}, \AppRef{1609}{1609}, \AppRef{1793}{1793}
\item Kayser, C.~G. \AppRef{1807a}{1807}
\item Kelland, P. \AppRef{1855}{1855}
\item Kepler, J. \AppRef{appendix}{}, \AppRef{1609}{1609}
\item Klein, F. \FnRef{103}
\item Klügel, G.~S. \FnRef{72}, \AppRef{1609}{1609}
\item Kullberg, J. \AppRef{1809}{1809}, \AppRef{1810}{1810}
\indexspace
\item La Frémoire, H.~C. de \AppRef{1852a}{1852}
\item Lagrange, J.~L. \AppRef{1609}{1609}
\item Lambert, J.~H. \AppRef{1772}{1772(?)}
\item Laplace, P.~S. \AppRef{1609}{1609}
\item Larmor, A. \AppRef{1901}{1901}
\item Laurent, P.~M.~H. \AppRef{1879}{1879}
\item Leeke, J. \FnRef{11}, \FnRef{30}, \FnRef{35}, \FnRef{50}
\item Leibnitz, G.~W. \AppRef{1687}{1687}
\item Leonardo Pisano, \textit{see} Pisano
\item Leslie, J. \AppRef{1821}{1821}
\item Leybourn, R. and W.  \FnRef{11}
\item L'Hospital, G.~F. de \AppRef{1687}{1687}
\item Lindman, C.~F. \AppRef{1870}{1870}
\item Loria, G.  \ParRef{2}, \FnRef{7}, \AppRef{1687}{1687}
\item Ludolph van Ceulen \AppRef{1615}{1615}
\item Lühmann, F. von \FnRef{111}
\indexspace
\item McDowell, J. \AppRef{1863}{1863}
\item Marinus \FnRef{11}
\item Matz, F.~P. \AppRef{1864a}{1864}
\item Mayer, J.~T. \AppRef{1783}{1783}
\item Menge, H. \FnRef{88}
\item Mersenne, M. \FnRef{111}
\item Mitzscherling, A.  \AppRef{appendix}{}
\item Mollweide, C.B. \FnRef{72}
% -----File: 100.png---Folio 88------------------------------------------------------
\item Montucla, J.~F.  \ParRef{19}, \FnRef{70}
\item Morville, N. \FnRef{67}, \AppRef{1793}{1793}
\item Muhammed Bagdedinus, \textit{see} el-Ba\.{g}\-dâdî
\item Muhammed b. `Abdelbâqî el-Ba\.{g}\-dâdî el-Fardî, \textit{see} el-Ba\.{g}\-dâdî
\item Muhammed b. Gâbir b. Sinân, Abû `Abdallâh el-Battânî, \textit{see} el-Battânî
\item Muhammed b. Muhammed el-Ba\.{g}\-dâdî, \textit{see} el-Ba\.{g}\-dâdî
\item Muhammed b. Muhammed el-Hasib Abû'l Wefâ, \textit{see} Abû'l Wefâ
\indexspace
\item Nemorarius, \textit{see} Jordanus
\item Newton, I.  \AppRef{appendix}{}, \AppRef{1676}{1676}, \linebreak[1] \AppRef{1747}{1747}
\item Nixon, R.~C.~J. \AppRef{1901}{1901}
\indexspace
\item Ofterdinger, L.F.  \ParRef{9}, \ParRef{15}, \FnRef{32}, \FnRef{38},
      \FnRef{51}, \FnRef{105}, \FnRef{111}
\item Ozanam, J. \AppRef{1688}{1688}, \AppRef{1699}{1699}
\indexspace
\item Pacioli, \textit{see} Paciuolo, L.
\item Paciuolo, L.  \ParRef{18}, \AppRef{appendix}{}
\item Palmer, R. \AppRef{1864a}{1864}
\item Pandya, N.~P. \AppRef{1772}{1772(?)}
\item Pappus of Alexandria  \ParRef{21}, \FnRef{85}, \FnRef{88},
      \FnRef{111}, \FnRef{113}
\item Paucker, G. \FnRef{111}
\item Pauly-Wissowa \FnRef{9}, \FnRef{81}
\item Perrin, E. \AppRef{1864a}{1864}
\item Pisano, Leonardo  \ParRef{10}, \ParRef{11}, \ParRef{12}, \ParRef{13},
      \ParRef{17}, \ParRef{18}, \ParRef{19}, \ParRef{22}, \ParRef{23},
      \ParRef{24}, \ParRef{25}, \ParRef{26}, \ParRef{27}, \ParRef{28},
      \ParRef{29}, \ParRef{30}, \ParRef{31}, \ParRef{32}, \ParRef{33},
      \ParRef{34}, \ParRef{35}, \ParRef{36}, \ParRef{37}, \ParRef{38},
      \ParRef{40}, \ParRef{41}, \ParRef{47}, \ParRef{48}, \ParRef{49},
      \ParRef{50}, \ParRef{51}, \ParRef{52}, \ParRef{53}, \ParRef{54},
      \ParRef{55}, \ParRef{56}, \ParRef{57},
      \FnRef{40}, \FnRef{41}, \FnRef{45}, \FnRef{46},
      \FnRef{47}, \FnRef{88}, \FnRef{96}, \FnRef{98},
      \FnRef{100}, \FnRef{107}, \FnRef{109}, \FnRef{110},
      \FnRef{111}, \FnRef{113},  \AppRef{appendix}{}
\item Planta, J.  \ParRef{4}, \FnRef{21}
\item Plato of Tivoli  \ParRef{18}
\item Potts, R. \AppRef{1847}{1847}
\item Proclus Diadochus  \ParRef{1}, \ParRef{6}, \ParRef{7}, \ParRef{18},
      \ParRef{49}, \FnRef{1}, \FnRef{2},
      \FnRef{35}, \FnRef{36}, \FnRef{103}, \AppRef{1560}{1560}
\item Ptolemy \FnRef{111}
\item Puissant, L. \AppRef{1801}{1801}
\indexspace
\item Quetelet, A. \FnRef{117}, \AppRef{1827}{1827}
\indexspace
\item Radhakrishnan  \AppRef{1864a}{1864}
\item Rebière, A.    \AppRef{1857}{1857}
\item Reinhold, E.   \AppRef{1574b}{1574}
\item Rhind Papyrus  \ParRef{20}
\item Richter, A. \FnRef{111}
\item Ritt, G.       \AppRef{1837}{1837}
\item Robison, J.    \AppRef{1855}{1855}
\item Rodham, J.     \AppRef{1795}{1795}
\item Ross, J.~A.    \AppRef{1805}{1805}
\item Rouché, E.     \AppRef{1687}{1687}
\item Rudd, T.       \AppRef{1612}{1612}
\item Rudio, F. \FnRef{116}
\item Rummer, F.     \AppRef{1852b}{1852}
\item Rutherford, W. \AppRef{1864a}{1864}
\indexspace
\item Saunderson, N. \AppRef{1740}{1740}
\item Savile, H. 6, \FnRef{34}
\item Schmid, W. \AppRef{1539}{1539}
\item Schoene, H. \FnRef{82}, \FnRef{83}
\item Schooten, F. van \AppRef{1657}{1657}
\item Schwenter, D. \AppRef{1667}{1667}
\item Sems, J. \AppRef{1612}{1612}
\item Serle, G. \FnRef{11}, \FnRef{30}, \FnRef{50}
\item Silberschlag, J. \AppRef{1772}{1772(?)}
\item Simplicius \FnRef{116}
\item Simpson, T. \AppRef{1609}{1609}, \AppRef{1747}{1747},
                  \AppRef{1752}{1752}, \AppRef{1902}{1902}
\item Simson, R.  \ParRef{39}, \FnRef{85}, \FnRef{88}, \FnRef{106},
      \FnRef{116}, \AppRef{1840}{1840}
\item Smith, C. \AppRef{1902}{1902}
\item Smith, T.  \ParRef{2}, \ParRef{3}, \ParRef{5}, \ParRef{19},
      \FnRef{14}, \FnRef{18}, \FnRef{66}
\item Snellius, W. \FnRef{111}, \AppRef{1615}{1615}
\item Speidell, J. \AppRef{1616}{1616}
\item Stanham, W.~C. \AppRef{1864a}{1864}
\item Steiner, J. \FnRef{117}
\item Steinschneider, M.  \ParRef{2}, \ParRef{3}, \ParRef{4}, \ParRef{6}
   \FnRef{12}, \FnRef{36}, \FnRef{53}, \FnRef{64}
\item Stevens, F.~H. \AppRef{1892}{1892}
\item Strode, T. \AppRef{1684}{1684}
\item Suter, H.  \ParRef{3}, \ParRef{4}, \ParRef{19}
   \FnRef{17}, \FnRef{36}, \FnRef{74},
   \FnRef{75}, \FnRef{76}, \FnRef{77}
\item Swinden, G.~H. van \FnRef{111}
\indexspace
\item \b{T}âbit b. Qorra  \ParRef{19}
\item Tartaglia, N. \AppRef{appendix}{}, \AppRef{1547a}{1547}, \AppRef{1560}{1560}
\item Taylor, C. \FnRef{103}
\item Tittel, K. \FnRef{81}
\item Townsend, R. \FnRef{117}
\indexspace
\item Urbin, Duke of  \ParRef{2}
\item Ussher, J.  \ParRef{2}, \ParRef{5}, \FnRef{15}
\indexspace
\item Vannson \FnRef{111}
\item Verhulst, P.~F. \AppRef{1827}{1827}
\item Viani de' Malatesti, F. \FnRef{11}
\item Vincent, A.~J.~C. \FnRef{83}
\indexspace
\item Wallis, J. \AppRef{1609}{1609}, \AppRef{1684}{1684}
\item Wenrich, J.~G.  \ParRef{2}, \FnRef{13}
\item Wiegand, A. \FnRef{111}
\item Wissowa \FnRef{9}, \FnRef{81}
\item Wölffing, E. \AppRef{1807b}{1807}, \AppRef{1844}{1844}
\item Woepcke, F.  \ParRef{7}, \ParRef{8}, \ParRef{9}, \ParRef{10},
      \ParRef{11}, \ParRef{12}, \ParRef{13}, \ParRef{18}, \ParRef{19}, \ParRef{57},
   \FnRef{36},  \FnRef{37},  \FnRef{46},  \FnRef{48},
   \FnRef{78},  \FnRef{79},  \FnRef{80},  \FnRef{86},
   \FnRef{101}, \FnRef{103}, \FnRef{104}, \FnRef{108},
   \FnRef{109}, \FnRef{111}, \FnRef{112}
\item Wright, J.~M.~F. \AppRef{1805}{1805}
\indexspace
\item Zdenek, R. \AppRef{1910b}{1910}
\item Zeuthen, H.G. \FnRef{103}
\item Zimmerman, L. \AppRef{1864b}{1864}
\renewcommand{\footrulewidth}{1pt}
\fancyfoot[C]{\small\textsc{cambridge: printed by john clay, m.a. at the university press}}
\end{theindex}
\pagestyle{empty}

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PDF statistics:
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 446 words of extra memory for PDF output out of 10000 (max. 10000000)

