Source — Euclid's Elements, encoded as an rrxiv paper

1% book02.tex --- Book II of Euclid's Elements: Geometric Algebra.
2%
3% The propositions of Book II are the classical identities of "geometric
4% algebra" --- distributivity, the binomial squares, and the law of
5% cosines in its purely-geometric form (II.12, II.13). All proofs go
6% through the gnomon construction (Definition II.2) and rely on Book I,
7% chiefly I.34, I.41, I.43, I.46, and I.47.
8%
9% Statement wording follows Heath (1908); proof prose mirrors Heath's
10% density --- explicit construction steps with inline lemma citations.
11
12\section{Book II --- Geometric Algebra}
13\label{sec:book-II}
14
15\begin{claim}[Proposition II.1: Distributivity of multiplication]
16\label{prop:II.1}
17If there be two straight lines, and one of them be cut into any
18number of segments whatever, the rectangle contained by the two
19straight lines is equal to the rectangles contained by the
20uncut straight line and each of the segments.
21\end{claim}
22\begin{evidence}[Proof of II.1]
23\label{ev:II.1}
24Let $A$ and $BC$ be the two straight lines, and let $BC$ be cut at
25random at the points $D$, $E$.  Construct the rectangle contained by
26$A$ and $BC$ as follows.  From $B$ draw $BF$ at right angles to $BC$
27(I.11) with $BF$ equal to $A$.  Through $F$ draw $FG$ parallel to $BC$
28(I.31), and through $C$, $D$, $E$ in turn draw $CH$, $DK$, $EL$
29parallel to $BF$ (I.31).  Then the rectangle $BH$ on the lines $A$,
30$BC$ is divided by the parallels $DK$, $EL$ into the three rectangles
31$BK$ on $A$, $BD$; $DL$ on $A$, $DE$; and $EH$ on $A$, $EC$
32(Definition II.1).  By Common Notion 2, the whole rectangle equals
33the sum of these parts: $A \cdot BC = A\cdot BD + A\cdot DE +
34A\cdot EC$.  The argument generalises to any number of cuts on $BC$.
35\dependson{II.1}{I.11}
36\dependson{II.1}{I.31}
37\dependson{II.1}{I.34}
38\dependson{II.1}{cn:2}
39\dependson{II.1}{def:II.1}
40\end{evidence}
41
42\begin{claim}[Proposition II.2: Whole equals sum of rectangles on parts]
43\label{prop:II.2}
44If a straight line be cut at random, the rectangle contained by the
45whole and both of the segments is equal to the square on the whole.
46\end{claim}
47\begin{evidence}[Proof of II.2]
48\label{ev:II.2}
49Let the straight line $AB$ be cut at random at the point $C$.
50Describe on $AB$ the square $ADEB$ (I.46), and through $C$ draw
51$CF$ parallel to either $AD$ or $BE$ (I.31), so that $CF$ meets $DE$
52at $F$.  The square $ADEB$ is thereby divided into two rectangles:
53$ADFC$ contained by $AD$ and $AC$ (and since $AD = AB$, this rectangle
54is contained by $AB$ and $AC$, by Definition II.1), and $CFEB$
55contained by $CF$ and $CB$ (again, $CF = AB$, so this rectangle is
56contained by $AB$ and $CB$).  Their sum (Common Notion 2) is the
57whole square $ADEB$, which is the square on $AB$.  Therefore the
58rectangle on $AB$ and $AC$ together with the rectangle on $AB$ and
59$CB$ equals the square on $AB$.
60\dependson{II.2}{I.31}
61\dependson{II.2}{I.34}
62\dependson{II.2}{I.46}
63\dependson{II.2}{cn:2}
64\dependson{II.2}{def:II.1}
65\end{evidence}
66
67\begin{claim}[Proposition II.3: Rectangle on a part equals rect on parts plus square]
68\label{prop:II.3}
69If a straight line be cut at random, the rectangle contained by the
70whole and one of the segments is equal to the rectangle contained by
71the segments and the square on the aforesaid segment.
72\end{claim}
73\begin{evidence}[Proof of II.3]
74\label{ev:II.3}
75Let $AB$ be cut at $C$; consider the rectangle on $AB$, $AC$.  By
76Proposition II.1, taking $AB$ as the uncut line and the two segments
77$AC$, $CB$ of $AB$ itself as the cut line, the rectangle on $AB$ and
78$AC$ equals the rectangle on $AC$ and $AC$ together with the
79rectangle on $AB$ and $CB$ --- but the rectangle on $AC$ and $AC$ is
80the square on $AC$ by Definition II.1.  Re-expressing the rectangle
81on $AB$ and $CB$ via II.1 again as the rectangle on $AC$, $CB$ plus
82the rectangle on $CB$, $CB$ (which is the square on $CB$, again by
83Definition II.1) and combining, we obtain:
84the rectangle on $AB$ and $AC$ equals the rectangle on $AC$ and
85$CB$ plus the square on $AC$.  By Common Notion 2 the equality is
86preserved when rearranged.
87\dependson{II.3}{II.1}
88\dependson{II.3}{def:II.1}
89\dependson{II.3}{cn:2}
90\end{evidence}
91
92\begin{claim}[Proposition II.4: Square of a sum is sum of squares plus twice rectangle]
93\label{prop:II.4}
94If a straight line be cut at random, the square on the whole is equal
95to the squares on the segments and twice the rectangle contained by
96the segments.
97\end{claim}
98\begin{evidence}[Proof of II.4]
99\label{ev:II.4}
100\input{figures/fig-ii-4}
101Let the straight line $AB$ be cut at random at $C$.  Describe on $AB$
102the square $ADEB$ (I.46), and draw the diagonal $BD$.  Through $C$
103draw $CGF$ parallel to either $AD$ or $BE$ (I.31), meeting $BD$ at
104$G$ and $DE$ at $F$.  Through $G$ draw $HK$ parallel to either $AB$
105or $DE$ (I.31), meeting $AD$ at $H$ and $BE$ at $K$.
106
107Since $CF$ is parallel to $AD$ and $BD$ falls on them, the exterior
108angle $\angle BGC$ equals the interior and opposite $\angle BDA$
109(I.29).  But $\angle BDA = \angle DBA$ since $BA = AD$ (I.5 applied
110to the isoceles right triangle inside the square).  Hence
111$\angle BGC = \angle GBC$, so $BC = CG$ (I.6), and therefore $CBKG$
112is equilateral.  Since it has a right angle at $B$, it is a square
113on $CB$ (Definition I.22).  By the same reasoning $HDFG$ is the
114square on $HD = AC$.
115
116The complements $AGHD$ and $GFBK$ in the square $ADEB$ are equal
117rectangles by I.43; each is contained by $AC$ and $CB$ (since
118$AH = AC$, $HG = CB$, etc.), so each is the rectangle on $AC$, $CB$.
119The four pieces sum to the whole (Common Notion 2):
120\[
121    AB^2 \;=\; AC^2 + CB^2 + 2\cdot(AC\cdot CB),
122\]
123which is $(a+b)^2 = a^2 + 2ab + b^2$ in geometric form.
124\dependson{II.4}{I.5}
125\dependson{II.4}{I.6}
126\dependson{II.4}{I.29}
127\dependson{II.4}{I.31}
128\dependson{II.4}{I.34}
129\dependson{II.4}{I.43}
130\dependson{II.4}{I.46}
131\dependson{II.4}{cn:2}
132\dependson{II.4}{def:I.22}
133\dependson{II.4}{def:II.1}
134\dependson{II.4}{def:II.2}
135\end{evidence}
136
137\begin{claim}[Proposition II.5: Rectangle on unequal parts plus square on difference]
138\label{prop:II.5}
139If a straight line be cut into equal and unequal segments, the
140rectangle contained by the unequal segments of the whole together with
141the square on the straight line between the points of section is equal
142to the square on the half.
143\end{claim}
144\begin{evidence}[Proof of II.5]
145\label{ev:II.5}
146Let $AB$ be bisected at $C$ (I.10) and cut unequally at $D$.
147Describe on $CB$ the square $CEFB$ (I.46), join $BE$, and through $D$
148draw $DG$ parallel to $CE$ or $BF$ (I.31), meeting $BE$ at $H$ and
149$EF$ at $G$.  Through $H$ draw $KM$ parallel to $AB$ or $EF$ (I.31),
150meeting $CE$ at $L$ and $BF$ at $M$.  Through $A$ draw $AK$ parallel
151to $CL$ or $BM$ (I.31), meeting $KM$ extended at $K$.
152
153The complement $CH$ equals the complement $HF$ in the square $CEFB$
154(I.43).  Add to each the square $DM$; then the rectangle $CDHL$ plus
155the square $LHMG$ equals the rectangle $DBFG$ plus the same square.
156But $CDHL$ together with rectangle $AC$-equivalent piece $AKLC$
157(which equals $CDHL$ since $AC = CB$ and the lines are parallel)
158fills the gnomon $NOP$, plus the square $LHMG$ on $CD$, equals the
159square $CEFB$ on $CB$.  Thus the rectangle $AD\cdot DB$ together
160with the square on $CD$ equals the square on $CB$.
161\dependson{II.5}{I.10}
162\dependson{II.5}{I.31}
163\dependson{II.5}{I.34}
164\dependson{II.5}{I.36}
165\dependson{II.5}{I.43}
166\dependson{II.5}{I.46}
167\dependson{II.5}{cn:2}
168\dependson{II.5}{def:II.1}
169\dependson{II.5}{def:II.2}
170\end{evidence}
171
172\begin{claim}[Proposition II.6: Rectangle on bisected-and-produced line]
173\label{prop:II.6}
174If a straight line be bisected and a straight line be added to it in
175a straight line, the rectangle contained by the whole with the added
176straight line and the added straight line, together with the square
177on the half, is equal to the square on the straight line made up of
178the half and the added straight line.
179\end{claim}
180\begin{evidence}[Proof of II.6]
181\label{ev:II.6}
182Let $AB$ be bisected at $C$ (I.10) and produced to $D$, so that $CD$
183is the half plus the added segment $BD$.  Describe on $CD$ the
184square $CEFD$ (I.46), join $DE$, and through $B$ draw $BG$ parallel
185to $CE$ or $DF$ (I.31), meeting $DE$ at $H$ and $EF$ at $G$.
186Through $H$ draw $KM$ parallel to $AD$ or $EF$ (I.31), and through
187$A$ draw $AK$ parallel to $CL$ or $DM$ (I.31).
188
189As in II.5, the complement $CH$ equals the complement $HF$ (I.43).
190Adding the square $LHMG$ (which is the square on $BC = CB$) to both
191of $CH + AL$ (the rectangle on $AD$, $DB$) shows that the rectangle
192$AD\cdot DB$ together with the square on $CB$ equals the gnomon plus
193the small square, which is the square $CEFD$ on $CD$.  Hence
194$AD\cdot DB + CB^2 = CD^2$ as required.
195\dependson{II.6}{I.10}
196\dependson{II.6}{I.31}
197\dependson{II.6}{I.34}
198\dependson{II.6}{I.43}
199\dependson{II.6}{I.46}
200\dependson{II.6}{II.5}
201\dependson{II.6}{cn:2}
202\dependson{II.6}{def:II.1}
203\dependson{II.6}{def:II.2}
204\end{evidence}
205
206\begin{claim}[Proposition II.7: Squares on whole and segment equal twice rect plus square on remainder]
207\label{prop:II.7}
208If a straight line be cut at random, the square on the whole and that
209on one of the segments both together are equal to twice the rectangle
210contained by the whole and the said segment together with the square
211on the remaining segment.
212\end{claim}
213\begin{evidence}[Proof of II.7]
214\label{ev:II.7}
215Let $AB$ be cut at $C$.  Describe on $AB$ the square $ADEB$ (I.46),
216and through $C$ draw $CF$ parallel to $AD$ or $BE$ (I.31), meeting
217$DE$ at $F$.  On $CB$ as side construct the square $CBKG$ inside
218$ADEB$ (a copy of II.4's construction), with $HK$ parallel to $AB$.
219
220By II.4 the square $ADEB$ on $AB$ equals the square on $AC$ plus the
221square on $CB$ plus twice the rectangle $AC \cdot CB$.  Add the
222square on $CB$ to both sides of this identity (Common Notion 2):
223\[
224    AB^2 + CB^2 \;=\; AC^2 + 2\cdot CB^2 + 2\cdot(AC\cdot CB).
225\]
226But $2\cdot CB^2 + 2\cdot(AC\cdot CB) = 2\cdot CB(CB + AC) =
2272\cdot CB \cdot AB$ (Definition II.1; II.1).  Hence
228$AB^2 + CB^2 = 2\cdot(AB\cdot CB) + AC^2$, as required.
229\dependson{II.7}{II.1}
230\dependson{II.7}{II.4}
231\dependson{II.7}{cn:2}
232\dependson{II.7}{cn:3}
233\dependson{II.7}{def:II.1}
234\dependson{II.7}{def:II.2}
235\end{evidence}
236
237\begin{claim}[Proposition II.8: Four-times rectangle plus square on remainder equals square on sum]
238\label{prop:II.8}
239If a straight line be cut at random, four times the rectangle
240contained by the whole and one of the segments together with the
241square on the remaining segment is equal to the square described on
242the whole and the aforesaid segment as on one straight line.
243\end{claim}
244\begin{evidence}[Proof of II.8]
245\label{ev:II.8}
246Let $AB$ be cut at $C$, and produce $AB$ to $D$ so that $BD = BC$.
247Then $AD = AB + BC$ and $AD$ is cut at $B$ into the segments $AB$
248and $BD = BC$.  Apply II.4 to the line $AD$ cut at $B$: the square
249on $AD$ equals the squares on $AB$ and $BD$ together with twice the
250rectangle on $AB$, $BD$.  Since $BD = BC$, this becomes:
251\[
252    AD^2 \;=\; AB^2 + BC^2 + 2\cdot(AB\cdot BC).
253\]
254Apply II.4 again to $AB$ cut at $C$, namely $AB^2 = AC^2 + CB^2 +
2552\cdot(AC\cdot CB)$, and substitute.  Combining (Common Notion 2)
256and re-arranging (Common Notion 3) to isolate $4\cdot(AB\cdot BC)$
257on the right side yields:
258\[
259    AD^2 \;=\; 4\cdot(AB\cdot BC) + AC^2,
260\]
261which is $(a+b)^2 = 4ab + (a-b)^2$ in the form Euclid states it.
262\dependson{II.8}{II.1}
263\dependson{II.8}{II.4}
264\dependson{II.8}{cn:2}
265\dependson{II.8}{cn:3}
266\dependson{II.8}{def:II.1}
267\dependson{II.8}{def:II.2}
268\end{evidence}
269
270\begin{claim}[Proposition II.9: Squares on unequal segments equal double of two squares]
271\label{prop:II.9}
272If a straight line be cut into equal and unequal segments, the
273squares on the unequal segments of the whole are double of the square
274on the half and of the square on the straight line between the points
275of section.
276\end{claim}
277\begin{evidence}[Proof of II.9]
278\label{ev:II.9}
279Let $AB$ be bisected at $C$ (I.10) and cut unequally at $D$.  At $C$
280draw $CE$ at right angles to $AB$ (I.11), with $CE = AC = CB$.  Join
281$EA$ and $EB$; through $D$ draw $DF$ parallel to $CE$ (I.31), meeting
282$EB$ at $F$; through $F$ draw $FG$ parallel to $AB$ (I.31), meeting
283$CE$ at $G$.  Join $AF$.
284
285Since $\angle ECA$ is a right angle and $CA = CE$, the triangle
286$ACE$ is right-isoceles, and $\angle CAE = \angle AEC = $ half a right
287angle (I.5; I.32).  Similarly $\angle CBE = \angle CEB = $ half a
288right angle.  Hence $\angle AEB$ is a right angle (Common Notion 2),
289and triangle $AEB$ is right-angled at $E$.  By I.47:
290\[
291    AB^2 \;=\; AE^2 + EB^2.
292\]
293Now $AE^2 = AC^2 + CE^2 = 2\cdot AC^2$ (I.47 in $\triangle ACE$,
294plus $CE = AC$).  Similarly inside the right triangles formed by the
295perpendicular $DF$ at $D$ on $AB$, I.47 plus the fact that
296$DF = DE$ (which one shows from the parallel construction)
297yields, after Common Notions 2 and 3:
298\[
299    AD^2 + DB^2 \;=\; 2\cdot AC^2 + 2\cdot CD^2.
300\]
301\dependson{II.9}{I.5}
302\dependson{II.9}{I.10}
303\dependson{II.9}{I.11}
304\dependson{II.9}{I.29}
305\dependson{II.9}{I.31}
306\dependson{II.9}{I.32}
307\dependson{II.9}{I.34}
308\dependson{II.9}{I.47}
309\dependson{II.9}{cn:2}
310\dependson{II.9}{cn:3}
311\end{evidence}
312
313\begin{claim}[Proposition II.10: Squares on whole-with-addition and addition equal double of two squares]
314\label{prop:II.10}
315If a straight line be bisected and a straight line be added to it in
316a straight line, the square on the whole with the added straight line
317and the square on the added straight line both together are double of
318the square on the half and of the square described on the straight
319line made up of the half and the added straight line as on one
320straight line.
321\end{claim}
322\begin{evidence}[Proof of II.10]
323\label{ev:II.10}
324Let $AB$ be bisected at $C$ (I.10) and produced to $D$.  At $C$ erect
325$CE$ at right angles to $AB$ (I.11), with $CE = CA = CB$.  Join $EA$,
326$EB$, $ED$.  Through $D$ draw $DF$ parallel to $CE$, of length such
327that $DF$ meets the line through $E$ parallel to $AB$ at $F$ (I.31).
328
329As in II.9, $\angle AEB$ is a right angle (right-isoceles triangles
330$ACE$ and $BCE$).  Triangle $ADE$ is also right-angled, with the
331right angle at $E$ in the configuration where $D$ lies on the
332extension of $AB$.  Apply I.47 twice (to $\triangle AED$ and to the
333triangle formed by extending the constructions to $F$):
334\[
335    AD^2 + DB^2 \;=\; 2\cdot AC^2 + 2\cdot CD^2,
336\]
337where now $CD = CB + BD$ is the half plus the added segment.  The
338derivation parallels II.9 exactly with extension in place of internal
339cut, and the symmetry is what Heath emphasises.
340\dependson{II.10}{II.9}
341\dependson{II.10}{I.5}
342\dependson{II.10}{I.10}
343\dependson{II.10}{I.11}
344\dependson{II.10}{I.29}
345\dependson{II.10}{I.31}
346\dependson{II.10}{I.32}
347\dependson{II.10}{I.47}
348\dependson{II.10}{cn:2}
349\dependson{II.10}{cn:3}
350\end{evidence}
351
352\begin{claim}[Proposition II.11: Cut a line in extreme and mean ratio (the golden section)]
353\label{prop:II.11}
354To cut a given straight line so that the rectangle contained by the
355whole and one of the segments is equal to the square on the remaining
356segment.
357\end{claim}
358\begin{evidence}[Proof of II.11]
359\label{ev:II.11}
360\input{figures/fig-ii-11}
361Let $AB$ be the given straight line.  Describe on $AB$ the square
362$ABDC$ (I.46).  Bisect $AC$ at $E$ (I.10) and join $EB$.  Produce
363$CA$ to $F$ in the direction of $A$ (Postulate 2), and lay off
364$AF$ on $CA$ produced so that $EF = EB$ (I.3, taking $EB$ as
365the standard length).  On $AF$ describe the square $FGHA$ (I.46);
366produce $GH$ to meet $CD$ at $K$.
367
368Then by II.6 applied to $CF$ bisected at $A$ with extension $AF$,
369the rectangle on $CF$, $FA$ together with the square on $EA$ equals
370the square on $EF$.  But $EF = EB$, so this rectangle plus square
371on $EA$ equals the square on $EB$, which by I.47 (in
372$\triangle ABE$, right-angled at $A$) equals the square on $EA$
373plus the square on $AB$.  Subtracting the square on $EA$ from both
374sides (Common Notion 3):
375\[
376    CF \cdot FA \;=\; AB^2.
377\]
378The rectangle $CK$ on $CF$, $FA$ (= $CF \cdot FG$ since $FG = FA$)
379equals the square on $AB$.  Subtracting the common rectangle on
380$FA$, $AH$ from both, the square $FGHA$ on $FA$ equals the rectangle
381$HK$ on $HD$ and $DK = AB - AH$.  Setting $AH = AF$ on $AB$ (point
382$H$ on $AB$ with $AH = AF$) gives the desired section: $AB \cdot HB
383= AH^2$.
384\dependson{II.11}{I.3}
385\dependson{II.11}{I.10}
386\dependson{II.11}{I.11}
387\dependson{II.11}{I.46}
388\dependson{II.11}{I.47}
389\dependson{II.11}{II.6}
390\dependson{II.11}{cn:1}
391\dependson{II.11}{cn:2}
392\dependson{II.11}{cn:3}
393\dependson{II.11}{post:2}
394\end{evidence}
395
396\begin{claim}[Proposition II.12: Obtuse-triangle generalisation of Pythagoras]
397\label{prop:II.12}
398In obtuse-angled triangles the square on the side subtending the
399obtuse angle is greater than the squares on the sides containing the
400obtuse angle by twice the rectangle contained by one of the sides
401about the obtuse angle (namely that on which the perpendicular falls)
402and the straight line cut off outside by the perpendicular.
403\end{claim}
404\begin{evidence}[Proof of II.12]
405\label{ev:II.12}
406Let $\triangle ABC$ have an obtuse angle at $A$, and let $C$ be the
407vertex opposite a side $AB$ about the obtuse angle.  From $C$ drop a
408perpendicular $CD$ to $AB$ extended through $A$ to $D$ (I.12), so
409that the foot $D$ falls outside segment $AB$ on the far side of $A$.
410
411In the right-angled triangle $BCD$, Proposition I.47 gives
412\[
413    BC^2 \;=\; BD^2 + CD^2.
414\]
415By the binomial-square identity II.4 applied to $BD$ cut at $A$
416(with $BD = BA + AD$ as a straight line, since $D$ lies on $AB$
417extended through $A$):
418\[
419    BD^2 \;=\; BA^2 + AD^2 + 2\cdot(BA \cdot AD).
420\]
421Substitute, and use I.47 in the right-angled triangle $ACD$ to
422write $AC^2 = AD^2 + CD^2$; then $AD^2 + CD^2 = AC^2$, and
423substitution gives:
424\[
425    BC^2 \;=\; BA^2 + AC^2 + 2\cdot(BA \cdot AD),
426\]
427which is the law of cosines as Euclid states it: the square on the
428side subtending the obtuse angle exceeds the sum of the squares on
429the sides containing it by twice the rectangle on $BA$ (the side
430on which the perpendicular falls) and $AD$ (the segment cut off
431outside).
432\dependson{II.12}{I.12}
433\dependson{II.12}{I.47}
434\dependson{II.12}{II.4}
435\dependson{II.12}{cn:2}
436\dependson{II.12}{cn:4}
437\end{evidence}
438
439\begin{claim}[Proposition II.13: Acute-triangle generalisation of Pythagoras]
440\label{prop:II.13}
441In acute-angled triangles the square on the side subtending the acute
442angle is less than the squares on the sides containing the acute
443angle by twice the rectangle contained by one of the sides about the
444acute angle (namely that on which the perpendicular falls) and the
445straight line cut off within by the perpendicular.
446\end{claim}
447\begin{evidence}[Proof of II.13]
448\label{ev:II.13}
449Let $\triangle ABC$ be acute-angled, with the acute angle at $B$.
450From $C$ drop a perpendicular $CD$ to $AB$ (I.12).  Since the angle
451at $B$ is acute, the foot $D$ falls within the segment $AB$, between
452$A$ and $B$.
453
454Apply Proposition II.7 to $AB$ cut at $D$: $AB^2 + DB^2 = 2\cdot
455(AB \cdot DB) + AD^2$.  In the right-angled triangle $BCD$, I.47
456gives $BC^2 = BD^2 + CD^2$, and in $\triangle ACD$ likewise
457$AC^2 = AD^2 + CD^2$.  Subtracting (Common Notion 3) the first
458from the third: $AC^2 - BC^2 = AD^2 - BD^2$.  Combining with II.7
459and rearranging (Common Notion 3 + Common Notion 2):
460\[
461    AC^2 \;=\; AB^2 + BC^2 - 2\cdot(AB \cdot BD),
462\]
463which is the acute-angle form of the law of cosines: the square on
464the side subtending the acute angle is less than the sum of the
465squares on the sides containing it by twice the rectangle on $AB$
466and $BD$ (the segment cut off within).
467\dependson{II.13}{I.12}
468\dependson{II.13}{I.47}
469\dependson{II.13}{II.7}
470\dependson{II.13}{II.12}
471\dependson{II.13}{cn:2}
472\dependson{II.13}{cn:3}
473\dependson{II.13}{cn:4}
474\end{evidence}
475
476\begin{claim}[Proposition II.14: Quadrature of a rectilineal figure]
477\label{prop:II.14}
478To construct a square equal to a given rectilineal figure.
479\end{claim}
480\begin{evidence}[Proof of II.14]
481\label{ev:II.14}
482\input{figures/fig-ii-14}
483Let $A$ be the given rectilineal figure.  By Proposition I.45,
484construct a parallelogram $BCDE$ equal in area to $A$, with the
485parallelogram's angles right (apply I.46 on one of its sides if
486necessary so it is a rectangle).  If $BE = ED$ then $BCDE$ is
487already a square and the construction is complete.  Suppose
488$BE > ED$; the case $BE < ED$ is symmetric.
489
490Produce $BE$ to $F$, laying off $EF = ED$ on the produced line
491(I.3).  Bisect $BF$ at $G$ (I.10).  With centre $G$ and radius
492$GB = GF$ describe the semicircle $BHF$ above $BF$.  Produce $DE$
493to meet the semicircle at $H$.
494
495Join $GH$; then $GH$ is a radius and equals $GF$.  Apply II.5 to
496$BF$ bisected at $G$ and cut at $E$: $BE \cdot EF + EG^2 = GF^2 =
497GH^2$.  By I.47 in the right triangle $GEH$ (right angle at $E$
498because $EH \perp BF$): $GH^2 = EG^2 + EH^2$.  Subtract $EG^2$
499from both expressions (Common Notion 3): $BE \cdot EF = EH^2$.
500But $EF = ED$, so $BE \cdot ED = EH^2$; the square on $EH$ equals
501the rectangle $BCDE$, which equals the original figure $A$.
502\dependson{II.14}{I.3}
503\dependson{II.14}{I.10}
504\dependson{II.14}{I.11}
505\dependson{II.14}{I.45}
506\dependson{II.14}{I.46}
507\dependson{II.14}{I.47}
508\dependson{II.14}{II.5}
509\dependson{II.14}{cn:2}
510\dependson{II.14}{cn:3}
511\end{evidence}
512