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

1% book01.tex — Book I of Euclid's Elements: Plane Geometry.
2%
3% Each Proposition X.Y is a `claim` with the statement; its proof is
4% an `evidence` block; the dependency DAG is encoded via \dependson
5% edges at the close of each proof.
6%
7% Statement wording follows Heath (1908), public domain, with minor
8% modernisation. Proof summaries are condensed for legibility; the
9% point of this encoding is to make the *dependency graph*
10% machine-readable, not to be a substitute for a full geometric exposition.
11
12\section{Book I --- Plane Geometry}
13\label{sec:book-I}
14
15\begin{claim}[Proposition I.1: Equilateral triangle on a given segment]
16\label{prop:I.1}
17On a given finite straight line to construct an equilateral triangle.
18\end{claim}
19
20\begin{evidence}[Proof of I.1]
21\label{ev:I.1}
22\input{figures/fig-i-1}
23Let $AB$ be the given finite straight line.  With centre $A$ and
24distance $AB$ describe the circle $BCD$ (Postulate 3).  With centre
25$B$ and distance $BA$ describe the circle $ACE$ (Postulate 3).  From
26the point $C$, where the circles cut one another, draw $CA$ and $CB$
27(Postulate 1).  Since $A$ is the centre of $BCD$, $AC = AB$
28(Definition I.15).  Since $B$ is the centre of $ACE$, $BC = BA$
29(Definition I.15).  By Common Notion 1, $AC = BC$.  Therefore the
30triangle $ABC$ is equilateral.
31\dependson{I.1}{post:1}
32\dependson{I.1}{post:3}
33\dependson{I.1}{def:I.15}
34\dependson{I.1}{cn:1}
35\end{evidence}
36
37\begin{claim}[Proposition I.2: Transfer a segment to a given point]
38\label{prop:I.2}
39To place at a given point (as an extremity) a straight line equal to a
40given straight line.
41\end{claim}
42\begin{evidence}[Proof of I.2]
43\label{ev:I.2}
44Apply I.1 to obtain an equilateral triangle; produce its sides
45(Postulate 2); describe a circle with centre at one endpoint of the
46given segment cutting one of the produced sides; the cut-off equals
47the given segment by Definition I.15 and Common Notion 1.
48\dependson{I.2}{I.1}
49\dependson{I.2}{post:1}
50\dependson{I.2}{post:2}
51\dependson{I.2}{post:3}
52\dependson{I.2}{def:I.15}
53\dependson{I.2}{cn:1}
54\end{evidence}
55
56\begin{claim}[Proposition I.3: Cut off a segment equal to a shorter]
57\label{prop:I.3}
58Given two unequal straight lines, to cut off from the greater a
59straight line equal to the less.
60\end{claim}
61\begin{evidence}[Proof of I.3]
62\label{ev:I.3}
63Use I.2 to construct, at one extremity of the longer line, a segment
64equal to the shorter.  Then by Definition I.15 the desired cut-off is
65obtained.
66\dependson{I.3}{I.2}
67\dependson{I.3}{def:I.15}
68\end{evidence}
69
70\begin{claim}[Proposition I.4: SAS congruence]
71\label{prop:I.4}
72If two triangles have two sides equal to two sides respectively, and
73have the angles contained by the equal straight lines equal, then they
74will also have the base equal to the base, the triangle will be equal
75to the triangle, and the remaining angles will be equal to the
76remaining angles respectively.
77\end{claim}
78\begin{evidence}[Proof of I.4]
79\label{ev:I.4}
80Superpose one triangle on the other; corresponding sides and angles
81coincide; by Common Notion 4 the figures are equal.
82\dependson{I.4}{cn:4}
83\end{evidence}
84
85\begin{claim}[Proposition I.5: Pons asinorum]
86\label{prop:I.5}
87In isosceles triangles the angles at the base are equal to one
88another; and if the equal straight lines be produced further, the
89angles under the base will be equal to one another.
90\end{claim}
91\begin{evidence}[Proof of I.5]
92\label{ev:I.5}
93\input{figures/fig-i-5}
94Apply I.3 to mark equal segments on the produced sides, then I.4 to
95two pairs of congruent triangles.  Common Notion 3 gives equality of
96the remaining angles.
97\dependson{I.5}{I.3}
98\dependson{I.5}{I.4}
99\dependson{I.5}{cn:3}
100\end{evidence}
101
102\begin{claim}[Proposition I.6: Converse of I.5]
103\label{prop:I.6}
104If in a triangle two angles are equal to one another, the sides which
105subtend the equal angles will also be equal to one another.
106\end{claim}
107\begin{evidence}[Proof of I.6]
108\label{ev:I.6}
109Suppose the sides unequal; cut off (by I.3) the greater equal to the
110less; by I.4 the resulting smaller triangle equals the whole, which
111contradicts Common Notion 5.  Therefore the sides are equal.
112\dependson{I.6}{I.3}
113\dependson{I.6}{I.4}
114\dependson{I.6}{cn:5}
115\end{evidence}
116
117\begin{claim}[Proposition I.7: Uniqueness of triangle on a base]
118\label{prop:I.7}
119Given two straight lines constructed on a straight line and meeting in
120a point, there cannot be constructed on the same straight line and on
121the same side of it two other straight lines meeting in another point
122and equal to the former two respectively, namely each to that which
123has the same extremity with it.
124\end{claim}
125\begin{evidence}[Proof of I.7]
126\label{ev:I.7}
127A double application of I.5 yields contradictory angle equalities; by
128Common Notion 5 the second meeting point cannot exist.
129\dependson{I.7}{I.5}
130\dependson{I.7}{cn:5}
131\end{evidence}
132
133\begin{claim}[Proposition I.8: SSS congruence]
134\label{prop:I.8}
135If two triangles have the two sides equal to two sides respectively,
136and have also the base equal to the base, they will also have the
137angles equal which are contained by the equal straight lines.
138\end{claim}
139\begin{evidence}[Proof of I.8]
140\label{ev:I.8}
141Superpose and apply I.7 to rule out a non-coincident image of the
142contained angle; conclude by Common Notion 4.
143\dependson{I.8}{I.7}
144\dependson{I.8}{cn:4}
145\end{evidence}
146
147\begin{claim}[Proposition I.9: Angle bisection]
148\label{prop:I.9}
149To bisect a given rectilineal angle.
150\end{claim}
151\begin{evidence}[Proof of I.9]
152\label{ev:I.9}
153Apply I.1 to construct an equilateral triangle on a chord between the
154angle's sides; the line from the vertex to the equilateral's apex
155bisects the angle by I.8.
156\dependson{I.9}{I.1}
157\dependson{I.9}{I.8}
158\end{evidence}
159
160\begin{claim}[Proposition I.10: Segment bisection]
161\label{prop:I.10}
162To bisect a given finite straight line.
163\end{claim}
164\begin{evidence}[Proof of I.10]
165\label{ev:I.10}
166Apply I.1 to erect an equilateral triangle on the segment; bisect the
167opposite angle by I.9; the bisector meets the segment at its midpoint
168by I.4.
169\dependson{I.10}{I.1}
170\dependson{I.10}{I.9}
171\dependson{I.10}{I.4}
172\end{evidence}
173
174\begin{claim}[Proposition I.11: Perpendicular at a point]
175\label{prop:I.11}
176To draw a straight line at right angles to a given straight line from
177a given point on it.
178\end{claim}
179\begin{evidence}[Proof of I.11]
180\label{ev:I.11}
181Apply I.3 to mark equal segments either side of the given point, then
182I.1 to erect an equilateral triangle whose apex line is perpendicular
183(by I.8 and Definition I.10).
184\dependson{I.11}{I.1}
185\dependson{I.11}{I.3}
186\dependson{I.11}{I.8}
187\dependson{I.11}{def:I.10}
188\end{evidence}
189
190\begin{claim}[Proposition I.12: Perpendicular from external point]
191\label{prop:I.12}
192To draw a perpendicular straight line to a given infinite straight
193line from a given point not on it.
194\end{claim}
195\begin{evidence}[Proof of I.12]
196\label{ev:I.12}
197With centre at the external point describe a circle (Postulate 3)
198cutting the given line; bisect the chord by I.10; the line from the
199external point to the midpoint is perpendicular by I.8.
200\dependson{I.12}{I.8}
201\dependson{I.12}{I.10}
202\dependson{I.12}{post:3}
203\end{evidence}
204
205\begin{claim}[Proposition I.13: Adjacent angles sum to two right angles]
206\label{prop:I.13}
207If a straight line set up on a straight line make angles, it will
208make either two right angles or angles equal to two right angles.
209\end{claim}
210\begin{evidence}[Proof of I.13]
211\label{ev:I.13}
212Drop a perpendicular by I.11; Common Notions 1--2 sum the resulting
213two right angles to the unequal-case angle sum.
214\dependson{I.13}{I.11}
215\dependson{I.13}{cn:1}
216\dependson{I.13}{cn:2}
217\dependson{I.13}{def:I.10}
218\end{evidence}
219
220\begin{claim}[Proposition I.14: Converse of I.13]
221\label{prop:I.14}
222If with any straight line, and at a point on it, two straight lines
223not lying on the same side make the adjacent angles equal to two
224right angles, the two straight lines will be in a straight line with
225one another.
226\end{claim}
227\begin{evidence}[Proof of I.14]
228\label{ev:I.14}
229Suppose not; then I.13 and Common Notion 1 yield contradictory angle
230sums.
231\dependson{I.14}{I.13}
232\dependson{I.14}{cn:1}
233\end{evidence}
234
235\begin{claim}[Proposition I.15: Vertical angles]
236\label{prop:I.15}
237If two straight lines cut one another, they make the vertical angles
238equal to one another.
239\end{claim}
240\begin{evidence}[Proof of I.15]
241\label{ev:I.15}
242Apply I.13 twice and Common Notion 3.
243\dependson{I.15}{I.13}
244\dependson{I.15}{cn:3}
245\end{evidence}
246
247\begin{claim}[Proposition I.16: Exterior angle of a triangle]
248\label{prop:I.16}
249In any triangle, if one of the sides be produced, the exterior angle
250is greater than either of the interior and opposite angles.
251\end{claim}
252\begin{evidence}[Proof of I.16]
253\label{ev:I.16}
254Bisect a side by I.10; produce a median; apply I.4 to obtain a
255congruent triangle that has the interior angle as one of its parts;
256Common Notion 5 closes the inequality.
257\dependson{I.16}{I.4}
258\dependson{I.16}{I.10}
259\dependson{I.16}{cn:5}
260\end{evidence}
261
262\begin{claim}[Proposition I.17: Sum of any two angles $<$ two right angles]
263\label{prop:I.17}
264In any triangle two angles taken together in any manner are less than
265two right angles.
266\end{claim}
267\begin{evidence}[Proof of I.17]
268\label{ev:I.17}
269Apply I.16 and I.13.
270\dependson{I.17}{I.13}
271\dependson{I.17}{I.16}
272\end{evidence}
273
274\begin{claim}[Proposition I.18: Greater side subtends greater angle]
275\label{prop:I.18}
276In any triangle the greater side subtends the greater angle.
277\end{claim}
278\begin{evidence}[Proof of I.18]
279\label{ev:I.18}
280Cut off (I.3) a segment on the greater side equal to the lesser;
281apply I.5 and I.16.
282\dependson{I.18}{I.3}
283\dependson{I.18}{I.5}
284\dependson{I.18}{I.16}
285\end{evidence}
286
287\begin{claim}[Proposition I.19: Converse of I.18]
288\label{prop:I.19}
289In any triangle the greater angle is subtended by the greater side.
290\end{claim}
291\begin{evidence}[Proof of I.19]
292\label{ev:I.19}
293Suppose not; by I.5 and I.18 contradiction.
294\dependson{I.19}{I.5}
295\dependson{I.19}{I.18}
296\end{evidence}
297
298\begin{claim}[Proposition I.20: Triangle inequality]
299\label{prop:I.20}
300In any triangle two sides taken together in any manner are greater
301than the remaining one.
302\end{claim}
303\begin{evidence}[Proof of I.20]
304\label{ev:I.20}
305Produce one side to an isosceles configuration by I.3; apply I.5 and
306I.19.
307\dependson{I.20}{I.3}
308\dependson{I.20}{I.5}
309\dependson{I.20}{I.19}
310\end{evidence}
311
312\begin{claim}[Proposition I.21: Interior cevian inequalities]
313\label{prop:I.21}
314If on one of the sides of a triangle, from its extremities, there be
315constructed two straight lines meeting within the triangle, the
316straight lines so constructed will be less than the remaining two
317sides of the triangle, but will contain a greater angle.
318\end{claim}
319\begin{evidence}[Proof of I.21]
320\label{ev:I.21}
321Two applications of I.20 and one of I.16.
322\dependson{I.21}{I.16}
323\dependson{I.21}{I.20}
324\end{evidence}
325
326\begin{claim}[Proposition I.22: Triangle from three given segments]
327\label{prop:I.22}
328Out of three straight lines, which are equal to three given straight
329lines, to construct a triangle: thus it is necessary that two of the
330straight lines taken together in any manner should be greater than the
331remaining one.
332\end{claim}
333\begin{evidence}[Proof of I.22]
334\label{ev:I.22}
335Apply I.20 (the necessity), then I.3 and Postulate 3 (the
336construction).
337\dependson{I.22}{I.3}
338\dependson{I.22}{I.20}
339\dependson{I.22}{post:3}
340\end{evidence}
341
342\begin{claim}[Proposition I.23: Reproduce a given angle]
343\label{prop:I.23}
344On a given straight line and at a point on it to construct a
345rectilineal angle equal to a given rectilineal angle.
346\end{claim}
347\begin{evidence}[Proof of I.23]
348\label{ev:I.23}
349Cut equal segments by I.3, construct the matching triangle by I.22,
350apply I.8.
351\dependson{I.23}{I.3}
352\dependson{I.23}{I.8}
353\dependson{I.23}{I.22}
354\end{evidence}
355
356\begin{claim}[Proposition I.24: Hinge inequality]
357\label{prop:I.24}
358If two triangles have the two sides equal to two sides respectively,
359but have the one of the angles contained by the equal straight lines
360greater than the other, they will also have the base greater than the
361base.
362\end{claim}
363\begin{evidence}[Proof of I.24]
364\label{ev:I.24}
365Apply I.23, I.4, I.5, and I.19.
366\dependson{I.24}{I.4}
367\dependson{I.24}{I.5}
368\dependson{I.24}{I.19}
369\dependson{I.24}{I.23}
370\end{evidence}
371
372\begin{claim}[Proposition I.25: Converse of I.24]
373\label{prop:I.25}
374If two triangles have the two sides equal to two sides respectively,
375but have the one base greater than the other, they will also have the
376one angle contained by the equal straight lines greater than the
377other.
378\end{claim}
379\begin{evidence}[Proof of I.25]
380\label{ev:I.25}
381Suppose not and use I.4, I.24.
382\dependson{I.25}{I.4}
383\dependson{I.25}{I.24}
384\end{evidence}
385
386\begin{claim}[Proposition I.26: ASA / AAS congruence]
387\label{prop:I.26}
388If two triangles have the two angles equal to two angles respectively,
389and one side equal to one side, namely either the side adjoining the
390equal angles or that subtending one of the equal angles, they will
391also have the remaining sides equal to the remaining sides and the
392remaining angle equal to the remaining angle.
393\end{claim}
394\begin{evidence}[Proof of I.26]
395\label{ev:I.26}
396Apply I.4 to the congruent angle--side--angle case; for the AAS case
397combine I.4 with I.16.
398\dependson{I.26}{I.4}
399\dependson{I.26}{I.16}
400\end{evidence}
401
402\begin{claim}[Proposition I.27: Alternate angles imply parallels]
403\label{prop:I.27}
404If a straight line falling on two straight lines make the alternate
405angles equal to one another, the straight lines will be parallel to
406one another.
407\end{claim}
408\begin{evidence}[Proof of I.27]
409\label{ev:I.27}
410Suppose the lines meet; I.16 gives a contradiction.  By Definition
411I.23 they are parallel.
412\dependson{I.27}{I.16}
413\dependson{I.27}{def:I.23}
414\end{evidence}
415
416\begin{claim}[Proposition I.28: Corresponding angles imply parallels]
417\label{prop:I.28}
418If a straight line falling on two straight lines make the exterior
419angle equal to the interior and opposite angle on the same side, or
420the interior angles on the same side equal to two right angles, the
421straight lines will be parallel to one another.
422\end{claim}
423\begin{evidence}[Proof of I.28]
424\label{ev:I.28}
425Reduce to I.27 via I.13 and I.15.
426\dependson{I.28}{I.13}
427\dependson{I.28}{I.15}
428\dependson{I.28}{I.27}
429\end{evidence}
430
431\begin{claim}[Proposition I.29: Properties of parallels (uses Postulate 5)]
432\label{prop:I.29}
433A straight line falling on parallel straight lines makes the alternate
434angles equal to one another, the exterior angle equal to the interior
435and opposite angle, and the interior angles on the same side equal to
436two right angles.
437\end{claim}
438\begin{evidence}[Proof of I.29]
439\label{ev:I.29}
440The first use of Postulate 5: any contrary supposition contradicts the
441parallel postulate via I.13.
442\dependson{I.29}{I.13}
443\dependson{I.29}{post:5}
444\end{evidence}
445
446\begin{claim}[Proposition I.30: Transitivity of parallelism]
447\label{prop:I.30}
448Straight lines parallel to the same straight line are also parallel
449to one another.
450\end{claim}
451\begin{evidence}[Proof of I.30]
452\label{ev:I.30}
453Two applications of I.29 and Common Notion 1.
454\dependson{I.30}{I.29}
455\dependson{I.30}{cn:1}
456\end{evidence}
457
458\begin{claim}[Proposition I.31: Construct a parallel through a point]
459\label{prop:I.31}
460Through a given point to draw a straight line parallel to a given
461straight line.
462\end{claim}
463\begin{evidence}[Proof of I.31]
464\label{ev:I.31}
465Apply I.23 to reproduce the alternate angle at the given point and
466I.27 to conclude parallelism.
467\dependson{I.31}{I.23}
468\dependson{I.31}{I.27}
469\end{evidence}
470
471\begin{claim}[Proposition I.32: Triangle angle sum]
472\label{prop:I.32}
473In any triangle, if one of the sides be produced, the exterior angle
474is equal to the two interior and opposite angles, and the three
475interior angles of the triangle are equal to two right angles.
476\end{claim}
477\begin{evidence}[Proof of I.32]
478\input{figures/fig-i-32}
479\label{ev:I.32}
480Apply I.31 to draw a parallel through the apex; apply I.29 twice and
481Common Notion 2.
482\dependson{I.32}{I.29}
483\dependson{I.32}{I.31}
484\dependson{I.32}{cn:2}
485\end{evidence}
486
487\begin{claim}[Proposition I.33: Connecting equal-and-parallel pairs]
488\label{prop:I.33}
489The straight lines joining equal and parallel straight lines (at the
490extremities which are in the same directions) are themselves equal
491and parallel.
492\end{claim}
493\begin{evidence}[Proof of I.33]
494\label{ev:I.33}
495Apply I.4 to the resulting two triangles and I.29 + I.27 for
496parallelism.
497\dependson{I.33}{I.4}
498\dependson{I.33}{I.27}
499\dependson{I.33}{I.29}
500\end{evidence}
501
502\begin{claim}[Proposition I.34: Properties of parallelograms]
503\label{prop:I.34}
504In parallelogrammic areas the opposite sides and angles are equal to
505one another, and the diameter bisects the areas.
506\end{claim}
507\begin{evidence}[Proof of I.34]
508\label{ev:I.34}
509Apply I.29 and I.26 to the two triangles cut by the diameter, then
510Common Notion 2.
511\dependson{I.34}{I.26}
512\dependson{I.34}{I.29}
513\dependson{I.34}{cn:2}
514\end{evidence}
515
516\begin{claim}[Proposition I.35: Parallelograms with same base \& between same parallels]
517\label{prop:I.35}
518Parallelograms which are on the same base and in the same parallels
519are equal to one another.
520\end{claim}
521\begin{evidence}[Proof of I.35]
522\label{ev:I.35}
523Apply I.34 and I.4 to congruent triangles; conclude by Common Notions
5242--3.
525\dependson{I.35}{I.4}
526\dependson{I.35}{I.34}
527\dependson{I.35}{cn:2}
528\dependson{I.35}{cn:3}
529\end{evidence}
530
531\begin{claim}[Proposition I.36: Parallelograms with equal bases]
532\label{prop:I.36}
533Parallelograms which are on equal bases and in the same parallels are
534equal to one another.
535\end{claim}
536\begin{evidence}[Proof of I.36]
537\label{ev:I.36}
538Translate one parallelogram via I.33 to share a base with the other,
539then apply I.35.
540\dependson{I.36}{I.33}
541\dependson{I.36}{I.35}
542\end{evidence}
543
544\begin{claim}[Proposition I.37: Triangles with same base \& parallels]
545\label{prop:I.37}
546Triangles which are on the same base and in the same parallels are
547equal to one another.
548\end{claim}
549\begin{evidence}[Proof of I.37]
550\label{ev:I.37}
551Complete each triangle to a parallelogram via I.31, then apply I.34
552and I.35.
553\dependson{I.37}{I.31}
554\dependson{I.37}{I.34}
555\dependson{I.37}{I.35}
556\end{evidence}
557
558\begin{claim}[Proposition I.38: Triangles with equal bases]
559\label{prop:I.38}
560Triangles which are on equal bases and in the same parallels are
561equal to one another.
562\end{claim}
563\begin{evidence}[Proof of I.38]
564\label{ev:I.38}
565Same construction as I.37 with I.36 in place of I.35.
566\dependson{I.38}{I.31}
567\dependson{I.38}{I.34}
568\dependson{I.38}{I.36}
569\end{evidence}
570
571\begin{claim}[Proposition I.39: Equal triangles on same base $\Rightarrow$ same parallels]
572\label{prop:I.39}
573Equal triangles which are on the same base and on the same side are
574also in the same parallels.
575\end{claim}
576\begin{evidence}[Proof of I.39]
577\label{ev:I.39}
578Apply I.31 to draw a parallel; I.37 forces the second vertex onto it.
579\dependson{I.39}{I.31}
580\dependson{I.39}{I.37}
581\end{evidence}
582
583\begin{claim}[Proposition I.40: Equal triangles on equal bases]
584\label{prop:I.40}
585Equal triangles which are on equal bases and on the same side are
586also in the same parallels.
587\end{claim}
588\begin{evidence}[Proof of I.40]
589\label{ev:I.40}
590Same approach as I.39 with I.38 in place of I.37.
591\dependson{I.40}{I.31}
592\dependson{I.40}{I.38}
593\end{evidence}
594
595\begin{claim}[Proposition I.41: Parallelogram is double a triangle]
596\label{prop:I.41}
597If a parallelogram have the same base with a triangle and be in the
598same parallels, the parallelogram is double of the triangle.
599\end{claim}
600\begin{evidence}[Proof of I.41]
601\label{ev:I.41}
602Apply I.34 and I.37; double via Common Notion 2.
603\dependson{I.41}{I.34}
604\dependson{I.41}{I.37}
605\dependson{I.41}{cn:2}
606\end{evidence}
607
608\begin{claim}[Proposition I.42: Construct a parallelogram equal in area to a triangle]
609\label{prop:I.42}
610To construct, in a given rectilineal angle, a parallelogram equal to
611a given triangle.
612\end{claim}
613\begin{evidence}[Proof of I.42]
614\label{ev:I.42}
615Bisect a side by I.10; apply I.23 to set the angle; apply I.31 and
616I.41.
617\dependson{I.42}{I.10}
618\dependson{I.42}{I.23}
619\dependson{I.42}{I.31}
620\dependson{I.42}{I.41}
621\end{evidence}
622
623\begin{claim}[Proposition I.43: Complements of a parallelogram]
624\label{prop:I.43}
625In any parallelogram the complements of the parallelograms about the
626diameter are equal to one another.
627\end{claim}
628\begin{evidence}[Proof of I.43]
629\label{ev:I.43}
630Apply I.34 to the bisecting diameter and Common Notions 2--3.
631\dependson{I.43}{I.34}
632\dependson{I.43}{cn:2}
633\dependson{I.43}{cn:3}
634\end{evidence}
635
636\begin{claim}[Proposition I.44: Apply a parallelogram to a segment equal to a triangle]
637\label{prop:I.44}
638To a given straight line to apply, in a given rectilineal angle, a
639parallelogram equal to a given triangle.
640\end{claim}
641\begin{evidence}[Proof of I.44]
642\label{ev:I.44}
643Apply I.42 and I.43 in sequence.
644\dependson{I.44}{I.42}
645\dependson{I.44}{I.43}
646\end{evidence}
647
648\begin{claim}[Proposition I.45: Apply a parallelogram equal to a rectilineal figure]
649\label{prop:I.45}
650To construct, in a given rectilineal angle, a parallelogram equal to
651a given rectilineal figure.
652\end{claim}
653\begin{evidence}[Proof of I.45]
654\label{ev:I.45}
655Triangulate the figure; sum the parallelograms by repeated I.44 and
656Common Notion 2.
657\dependson{I.45}{I.44}
658\dependson{I.45}{cn:2}
659\end{evidence}
660
661\begin{claim}[Proposition I.46: Construct a square on a segment]
662\label{prop:I.46}
663On a given straight line to describe a square.
664\end{claim}
665\begin{evidence}[Proof of I.46]
666\label{ev:I.46}
667Apply I.11 to erect perpendiculars; use I.3 to cut off equal segments;
668use I.31 and I.29 to close the square; verify right angles via I.34.
669\dependson{I.46}{I.3}
670\dependson{I.46}{I.11}
671\dependson{I.46}{I.29}
672\dependson{I.46}{I.31}
673\dependson{I.46}{I.34}
674\end{evidence}
675
676\begin{claim}[Proposition I.47: Pythagoras' theorem]
677\label{prop:I.47}
678In right-angled triangles the square on the side subtending the right
679angle is equal to the squares on the sides containing the right angle.
680\end{claim}
681\begin{evidence}[Proof of I.47]
682\input{figures/fig-i-47}
683\label{ev:I.47}
684Erect squares on each side via I.46; using I.14, I.31 and I.41 show
685each part-square equals a corresponding parallelogram cut off the
686hypotenuse-square by the perpendicular from the right angle; sum via
687Common Notion 2.
688\dependson{I.47}{I.4}
689\dependson{I.47}{I.14}
690\dependson{I.47}{I.31}
691\dependson{I.47}{I.41}
692\dependson{I.47}{I.46}
693\dependson{I.47}{cn:2}
694\supports{I.47}{I.48}
695\end{evidence}
696
697\begin{claim}[Proposition I.48: Converse of Pythagoras]
698\label{prop:I.48}
699If in a triangle the square on one of the sides be equal to the
700squares on the remaining two sides of the triangle, the angle
701contained by the remaining two sides of the triangle is right.
702\end{claim}
703\begin{evidence}[Proof of I.48]
704\label{ev:I.48}
705Construct a right triangle with the same two legs by I.11; apply
706I.47, I.8, and Common Notion 1.
707\dependson{I.48}{I.8}
708\dependson{I.48}{I.11}
709\dependson{I.48}{I.47}
710\dependson{I.48}{cn:1}
711\end{evidence}
712