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

1% book11.tex --- Book XI of Euclid's Elements: Solid Geometry.
2%
3% All 39 propositions encoded.  Book XI extends the plane geometry of
4% Books I-IV to three dimensions: lines and planes in space, parallel
5% planes, dihedral angles, parallelepipeds, and the foundational lemmas
6% for the Platonic solids of Book XIII.
7%
8% Wording follows Heath (1908).
9
10\section{Book XI --- Solid Geometry}
11\label{sec:book-XI}
12
13\begin{claim}[Proposition XI.1: A straight line cannot have part in a plane and part not]
14\label{prop:XI.1}
15A part of a straight line cannot be in the plane of reference and a
16part in a plane more elevated.
17\end{claim}
18\begin{evidence}[Proof of XI.1]
19\label{ev:XI.1}
20If a straight line $AB$ had part in one plane and continued part in
21another, then through $B$ there would be two distinct straight lines
22from $A$ (one in each plane), contradicting Postulate 1 (uniqueness
23of the straight line through two points).
24\dependson{XI.1}{post:1}
25\dependson{XI.1}{def:I.4}
26\end{evidence}
27
28\begin{claim}[Proposition XI.2: Intersecting lines determine a plane]
29\label{prop:XI.2}
30If two straight lines cut one another, they are in one plane, and
31every triangle is in one plane.
32\end{claim}
33\begin{evidence}[Proof of XI.2]
34\label{ev:XI.2}
35Two intersecting lines pick out three non-collinear points (one at
36the intersection, one on each line); through these three points
37passes exactly one plane (the analogue in 3D of Postulate 1).
38\dependson{XI.2}{XI.1}
39\dependson{XI.2}{def:I.7}
40\end{evidence}
41
42\begin{claim}[Proposition XI.3: Intersection of two planes is a straight line]
43\label{prop:XI.3}
44If two planes cut one another, their common section is a straight
45line.
46\end{claim}
47\begin{evidence}[Proof of XI.3]
48\label{ev:XI.3}
49Take two points $A$, $B$ in the common section.  Draw the straight
50line $AB$ in each plane; by XI.1 each segment of $AB$ lies in its
51plane, and uniqueness of the line forces both segments to coincide.
52\dependson{XI.3}{XI.1}
53\dependson{XI.3}{post:1}
54\end{evidence}
55
56\begin{claim}[Proposition XI.4: Line perpendicular to two intersecting lines is perpendicular to their plane]
57\label{prop:XI.4}
58If a straight line be set up at right angles to two straight lines
59which cut one another, at their common point of section, it will
60also be at right angles to the plane through them.
61\end{claim}
62\begin{evidence}[Proof of XI.4]
63\label{ev:XI.4}
64Take any other straight line $\ell$ through the foot in the plane;
65$\ell$ can be expressed as a sum of perpendicular components on the
66two given lines (by I.46-style decomposition), and the perpendicular
67to both is perpendicular to the sum by I.4 applied to the right
68triangles formed.
69\dependson{XI.4}{I.4}
70\dependson{XI.4}{I.8}
71\dependson{XI.4}{def:XI.3}
72\end{evidence}
73
74\begin{claim}[Proposition XI.5: Three lines through a point each perpendicular to a fourth lie in a plane]
75\label{prop:XI.5}
76If a straight line be set up at right angles to three straight lines
77which meet one another, at their common point of section, the three
78straight lines are in one plane.
79\end{claim}
80\begin{evidence}[Proof of XI.5]
81\label{ev:XI.5}
82Two of the three meeting lines determine a plane (XI.2); if the
83third were out of that plane, the perpendicular relation combined
84with XI.4 would force two distinct planes through the same set of
85perpendicular lines, contradiction.
86\dependson{XI.5}{XI.2}
87\dependson{XI.5}{XI.4}
88\end{evidence}
89
90\begin{claim}[Proposition XI.6: Two lines perpendicular to the same plane are parallel]
91\label{prop:XI.6}
92If two straight lines be at right angles to the same plane, the
93straight lines will be parallel.
94\end{claim}
95\begin{evidence}[Proof of XI.6]
96\label{ev:XI.6}
97Suppose the two perpendiculars $AB$, $CD$ met or were skew.  Drop
98the segment $BD$ in the plane; the angles at $B$ and $D$ are right.
99In the plane through $AB$ and $BD$, by I.28 the line $CD$ would
100need to be parallel to $AB$, fixing the plane through $AB$, $CD$.
101Then within that plane the two right angles force $AB \parallel CD$.
102\dependson{XI.6}{I.28}
103\dependson{XI.6}{XI.2}
104\dependson{XI.6}{XI.3}
105\dependson{XI.6}{XI.4}
106\end{evidence}
107
108\begin{claim}[Proposition XI.7: A line in a plane parallel to a parallel line in another plane lies in the connecting plane]
109\label{prop:XI.7}
110If two straight lines be parallel, and points be taken at random on
111each of them, the straight line joining the points is in the same
112plane with the parallels.
113\end{claim}
114\begin{evidence}[Proof of XI.7]
115\label{ev:XI.7}
116Two parallel lines determine a plane (XI.2 extended).  Any joining
117segment between points on the two parallels lies in this plane by
118XI.1.
119\dependson{XI.7}{XI.1}
120\dependson{XI.7}{XI.2}
121\end{evidence}
122
123\begin{claim}[Proposition XI.8: Two parallel lines, one perpendicular to a plane, the other also perpendicular]
124\label{prop:XI.8}
125If two straight lines be parallel, and one of them be at right angles
126to any plane, the remaining one will also be at right angles to the
127same plane.
128\end{claim}
129\begin{evidence}[Proof of XI.8]
130\label{ev:XI.8}
131By XI.7 the connecting segment lies in the plane of the parallels;
132combine XI.4 (perpendicularity transferred along parallel lines via
133shared plane structure) with I.29 (alternate angles) to obtain
134perpendicularity of the second line.
135\dependson{XI.8}{I.29}
136\dependson{XI.8}{XI.4}
137\dependson{XI.8}{XI.6}
138\dependson{XI.8}{XI.7}
139\end{evidence}
140
141\begin{claim}[Proposition XI.9: Lines parallel to the same line are parallel]
142\label{prop:XI.9}
143Straight lines which are parallel to the same straight line and are
144not in the same plane with it are also parallel to one another.
145\end{claim}
146\begin{evidence}[Proof of XI.9]
147\label{ev:XI.9}
148Construct in each plane perpendiculars from a common point to the
149shared straight line; the perpendiculars are equal in length, and
150by I.33 the resulting transversal is parallel to both targets.
151\dependson{XI.9}{I.33}
152\dependson{XI.9}{XI.6}
153\dependson{XI.9}{XI.8}
154\end{evidence}
155
156\begin{claim}[Proposition XI.10: Two angles with parallel sides are equal in space]
157\label{prop:XI.10}
158If two straight lines meeting one another be parallel to two straight
159lines meeting one another, not in the same plane, they will contain
160equal angles.
161\end{claim}
162\begin{evidence}[Proof of XI.10]
163\label{ev:XI.10}
164Construct the parallelogram joining corresponding points; by I.33
165opposite sides are equal, and by I.8 (SSS) the two triangles formed
166at the vertex angles are congruent.
167\dependson{XI.10}{I.8}
168\dependson{XI.10}{I.33}
169\dependson{XI.10}{XI.9}
170\end{evidence}
171
172\begin{claim}[Proposition XI.11: Drop a perpendicular from an external point to a plane]
173\label{prop:XI.11}
174From a given elevated point to draw a straight line perpendicular to
175a given plane.
176\end{claim}
177\begin{evidence}[Proof of XI.11]
178\label{ev:XI.11}
179Drop a chord through the point parallel to the plane; drop a
180perpendicular from the chord to its foot in the plane; the
181constructed line, being perpendicular to two intersecting lines at
182the foot, is perpendicular to the plane (XI.4).
183\dependson{XI.11}{I.11}
184\dependson{XI.11}{I.12}
185\dependson{XI.11}{XI.4}
186\end{evidence}
187
188\begin{claim}[Proposition XI.12: Erect a perpendicular to a plane at a given point in the plane]
189\label{prop:XI.12}
190To set up a straight line at right angles to a given plane from a
191given point in it.
192\end{claim}
193\begin{evidence}[Proof of XI.12]
194\label{ev:XI.12}
195Take an external point above the plane; drop a perpendicular from it
196to the plane via XI.11; the foot may be made to coincide with the
197given point by an additional parallel translation (using XI.8).
198\dependson{XI.12}{XI.8}
199\dependson{XI.12}{XI.11}
200\end{evidence}
201
202\begin{claim}[Proposition XI.13: A unique perpendicular to a plane from a given point]
203\label{prop:XI.13}
204From the same point two straight lines cannot be set up at right
205angles to the same plane on the same side.
206\end{claim}
207\begin{evidence}[Proof of XI.13]
208\label{ev:XI.13}
209Two such perpendiculars would meet two lines in the plane at the
210same right angles; by XI.4 / XI.6 the two perpendiculars would have
211to be parallel; but parallels do not meet -- contradicting their
212common origin.
213\dependson{XI.13}{XI.4}
214\dependson{XI.13}{XI.6}
215\end{evidence}
216
217\begin{claim}[Proposition XI.14: Planes perpendicular to the same line are parallel]
218\label{prop:XI.14}
219Planes to which the same straight line is at right angles will be
220parallel.
221\end{claim}
222\begin{evidence}[Proof of XI.14]
223\label{ev:XI.14}
224If the two planes met, the line of intersection (XI.3) would meet
225the common perpendicular at right angles in two distinct places --
226contradicting XI.13.
227\dependson{XI.14}{XI.3}
228\dependson{XI.14}{XI.13}
229\dependson{XI.14}{def:XI.8}
230\end{evidence}
231
232\begin{claim}[Proposition XI.15: Two intersecting line-pairs parallel to two other line-pairs give parallel planes]
233\label{prop:XI.15}
234If two straight lines meeting one another be parallel to two straight
235lines meeting one another, not being in the same plane, the planes
236through them are parallel.
237\end{claim}
238\begin{evidence}[Proof of XI.15]
239\label{ev:XI.15}
240By XI.10 the angles are equal; drop a common perpendicular line; by
241XI.14 the two planes share a common perpendicular and are parallel.
242\dependson{XI.15}{XI.9}
243\dependson{XI.15}{XI.10}
244\dependson{XI.15}{XI.14}
245\end{evidence}
246
247\begin{claim}[Proposition XI.16: A plane cuts parallel planes in parallel lines]
248\label{prop:XI.16}
249If two parallel planes be cut by any plane, their common sections
250are parallel.
251\end{claim}
252\begin{evidence}[Proof of XI.16]
253\label{ev:XI.16}
254The two intersection lines lie in the cutting plane, and if they
255met, the meeting point would belong to both parallel planes -- which
256is impossible.
257\dependson{XI.16}{XI.3}
258\dependson{XI.16}{def:XI.8}
259\end{evidence}
260
261\begin{claim}[Proposition XI.17: Two parallel planes cut a transversal proportionally]
262\label{prop:XI.17}
263If two straight lines be cut by parallel planes, they will be cut in
264the same ratios.
265\end{claim}
266\begin{evidence}[Proof of XI.17]
267\label{ev:XI.17}
268Draw a parallelogram structure between the parallel planes; apply
269VI.2 (basic proportionality) in each pair of cross-sectional lines.
270\dependson{XI.17}{VI.2}
271\dependson{XI.17}{XI.16}
272\end{evidence}
273
274\begin{claim}[Proposition XI.18: A line perpendicular to a plane makes the plane through it perpendicular to that plane]
275\label{prop:XI.18}
276If a straight line be at right angles to any plane, all the planes
277through it will also be at right angles to the same plane.
278\end{claim}
279\begin{evidence}[Proof of XI.18]
280\label{ev:XI.18}
281Any plane through the perpendicular contains the perpendicular line;
282by Definition XI.4 (perpendicular planes) the containing plane is
283perpendicular to the original plane.
284\dependson{XI.18}{XI.4}
285\dependson{XI.18}{def:XI.4}
286\end{evidence}
287
288\begin{claim}[Proposition XI.19: Two perpendicular planes intersect in a line perpendicular to the base]
289\label{prop:XI.19}
290If two planes which cut one another be at right angles to any plane,
291their common section will also be at right angles to the same plane.
292\end{claim}
293\begin{evidence}[Proof of XI.19]
294\label{ev:XI.19}
295The intersection line lies in both planes; by Definition XI.4 the
296perpendiculars from any point of the intersection within each plane
297are perpendicular to the base plane; XI.13 then forces the
298intersection line itself to be perpendicular.
299\dependson{XI.19}{XI.13}
300\dependson{XI.19}{XI.18}
301\dependson{XI.19}{def:XI.4}
302\end{evidence}
303
304\begin{claim}[Proposition XI.20: Solid angle inequality (triangle inequality for face-angles)]
305\label{prop:XI.20}
306If a solid angle be contained by three plane angles, any two, taken
307together in any manner, are greater than the remaining one.
308\end{claim}
309\begin{evidence}[Proof of XI.20]
310\label{ev:XI.20}
311Suppose the largest face-angle is $\angle BAC$.  Within $\angle BAC$
312construct $\angle BAD$ equal to $\angle BAE$ (one of the other
313face-angles).  By I.4 / I.24, the corresponding chord arcs in space
314give the desired strict triangle-style inequality among the
315face-angles.
316\dependson{XI.20}{I.4}
317\dependson{XI.20}{I.20}
318\dependson{XI.20}{I.24}
319\end{evidence}
320
321\begin{claim}[Proposition XI.21: Sum of face-angles at a solid angle is less than four right angles]
322\label{prop:XI.21}
323Any solid angle is contained by plane angles less than four right
324angles.
325\end{claim}
326\begin{evidence}[Proof of XI.21]
327\label{ev:XI.21}
328Cut a small polygon by a plane near the apex; the sum of the
329exterior angles of this polygon is less than $4 \cdot 90^\circ$ (by
330I.32 / I.34 applied to the polygon).  The interior face-angles at
331the apex are the supplements of these exterior angles, so their sum
332falls strictly short of $4 \cdot 90^\circ$.
333\dependson{XI.21}{I.32}
334\dependson{XI.21}{XI.20}
335\end{evidence}
336
337\begin{claim}[Proposition XI.22: Three face-angles whose sum is less than four right angles can form a solid angle]
338\label{prop:XI.22}
339If there be three plane angles of which two, taken together in any
340manner, are greater than the remaining one, and they are contained
341by equal straight lines, it is possible to construct a triangle out
342of the straight lines joining the extremities of the equal straight
343lines.
344\end{claim}
345\begin{evidence}[Proof of XI.22]
346\label{ev:XI.22}
347The plane-angle inequality (XI.20) is precisely the triangle
348inequality for the joining chords; by I.22 (construction of a
349triangle on three given segments) the resulting triangle exists.
350\dependson{XI.22}{I.22}
351\dependson{XI.22}{XI.20}
352\end{evidence}
353
354\begin{claim}[Proposition XI.23: Construct a solid angle from three given face-angles]
355\label{prop:XI.23}
356To construct a solid angle out of three plane angles, two of which,
357taken together in any manner, are greater than the remaining one;
358thus the sum of the three angles must be less than four right angles.
359\end{claim}
360\begin{evidence}[Proof of XI.23]
361\label{ev:XI.23}
362Apply XI.22 to obtain the triangle of chords; mount that triangle so
363that the three face-angles meet at a common apex; the bound of XI.21
364ensures consistency.
365\dependson{XI.23}{XI.20}
366\dependson{XI.23}{XI.21}
367\dependson{XI.23}{XI.22}
368\end{evidence}
369
370\begin{claim}[Proposition XI.24: Parallelepiped has parallelogram faces]
371\label{prop:XI.24}
372If a solid be contained by parallel planes, the opposite planes in
373it are equal and similar parallelograms.
374\end{claim}
375\begin{evidence}[Proof of XI.24]
376\label{ev:XI.24}
377Opposite faces share parallel sides (by XI.16) and equal angles (by
378XI.10), so they are congruent parallelograms by I.33 / I.34.
379\dependson{XI.24}{I.33}
380\dependson{XI.24}{I.34}
381\dependson{XI.24}{XI.10}
382\dependson{XI.24}{XI.16}
383\end{evidence}
384
385\begin{claim}[Proposition XI.25: Parallelepiped cut by a plane parallel to a face is divided proportionally]
386\label{prop:XI.25}
387If a parallelepipedal solid be cut by a plane parallel to opposite
388planes, then, as the base is to the base, so will the solid be to
389the solid.
390\end{claim}
391\begin{evidence}[Proof of XI.25]
392\label{ev:XI.25}
393Apply XI.17 to the side faces and VI.1 to the parallel base-pairs;
394the volume is proportional to one varying side at constant
395cross-section.
396\dependson{XI.25}{VI.1}
397\dependson{XI.25}{XI.17}
398\dependson{XI.25}{XI.24}
399\end{evidence}
400
401\begin{claim}[Proposition XI.26: Construct a solid angle equal to a given solid angle]
402\label{prop:XI.26}
403At a given point on a given straight line to construct a solid angle
404equal to a given solid angle contained by three plane angles.
405\end{claim}
406\begin{evidence}[Proof of XI.26]
407\label{ev:XI.26}
408Reproduce each face-angle by I.23 in the appropriate planes; by XI.23
409the resulting figure determines a solid angle congruent to the given
410one.
411\dependson{XI.26}{I.23}
412\dependson{XI.26}{XI.10}
413\dependson{XI.26}{XI.23}
414\end{evidence}
415
416\begin{claim}[Proposition XI.27: Construct a parallelepiped similar to a given parallelepiped on a given edge]
417\label{prop:XI.27}
418On a given straight line to construct a parallelepipedal solid
419similar and similarly situated to a given parallelepipedal solid.
420\end{claim}
421\begin{evidence}[Proof of XI.27]
422\label{ev:XI.27}
423Apply the face-angle construction of XI.26 at each vertex; by VI.18
424the resulting faces are similar to the corresponding faces of the
425given solid.
426\dependson{XI.27}{VI.18}
427\dependson{XI.27}{XI.24}
428\dependson{XI.27}{XI.26}
429\end{evidence}
430
431\begin{claim}[Proposition XI.28: Parallelepiped bisected by its diagonal plane]
432\label{prop:XI.28}
433If a parallelepipedal solid be cut by a plane through the diagonals
434of the opposite planes, the solid will be bisected by the plane.
435\end{claim}
436\begin{evidence}[Proof of XI.28]
437\label{ev:XI.28}
438The two pieces are mirror-image prisms with congruent base-triangles
439(by I.34); by XI.24 their volumes are equal.
440\dependson{XI.28}{I.34}
441\dependson{XI.28}{XI.24}
442\dependson{XI.28}{XI.25}
443\end{evidence}
444
445\begin{claim}[Proposition XI.29: Parallelepipeds on equal bases with same height are equal]
446\label{prop:XI.29}
447Parallelepipedal solids which are on the same base and of the same
448height, and in which the extremities of the sides which stand up are
449on the same straight lines, are equal to one another.
450\end{claim}
451\begin{evidence}[Proof of XI.29]
452\label{ev:XI.29}
453The two parallelepipeds can be decomposed into congruent prisms via
454XI.28 and I.34 applied repeatedly; the equality is the 3D analogue
455of I.35 (parallelograms on the same base between the same parallels).
456\dependson{XI.29}{I.35}
457\dependson{XI.29}{XI.24}
458\dependson{XI.29}{XI.28}
459\end{evidence}
460
461\begin{claim}[Proposition XI.30: Parallelepipeds on equal bases with same height but different oblique placement are equal]
462\label{prop:XI.30}
463Parallelepipedal solids which are on the same base and of the same
464height, and in which the extremities of the sides which stand up are
465not on the same straight lines, are equal to one another.
466\end{claim}
467\begin{evidence}[Proof of XI.30]
468\label{ev:XI.30}
469Variant of XI.29 with the oblique sides at different angles; the
470decomposition argument still applies after a shear.
471\dependson{XI.30}{XI.29}
472\end{evidence}
473
474\begin{claim}[Proposition XI.31: Parallelepipeds on equal bases are in the ratio of their heights]
475\label{prop:XI.31}
476Parallelepipedal solids which are on equal bases and of the same
477height are equal to one another.
478\end{claim}
479\begin{evidence}[Proof of XI.31]
480\label{ev:XI.31}
481By XI.25 the volume is proportional to the base at fixed height; if
482the bases are equal, the volumes are equal.
483\dependson{XI.31}{XI.25}
484\dependson{XI.31}{XI.29}
485\dependson{XI.31}{XI.30}
486\end{evidence}
487
488\begin{claim}[Proposition XI.32: Parallelepipeds of equal height are as their bases]
489\label{prop:XI.32}
490Parallelepipedal solids which are of the same height are to one
491another as their bases.
492\end{claim}
493\begin{evidence}[Proof of XI.32]
494\label{ev:XI.32}
495By XI.25 applied with one shared dimension fixed.
496\dependson{XI.32}{XI.25}
497\dependson{XI.32}{XI.31}
498\end{evidence}
499
500\begin{claim}[Proposition XI.33: Similar parallelepipeds are in the triplicate ratio of corresponding edges]
501\label{prop:XI.33}
502Similar parallelepipedal solids are to one another in the triplicate
503ratio of their corresponding sides.
504\end{claim}
505\begin{evidence}[Proof of XI.33]
506\label{ev:XI.33}
507The 3D analogue of VI.20: scaling each of the three edges by ratio
508$k$ multiplies the volume by $k^3$.  Apply VI.20 to two faces and
509XI.32 to extrude.
510\dependson{XI.33}{VI.20}
511\dependson{XI.33}{XI.32}
512\dependson{XI.33}{def:V.10}
513\end{evidence}
514
515\begin{claim}[Proposition XI.34: Equal parallelepipeds have reciprocally proportional edges]
516\label{prop:XI.34}
517In equal parallelepipedal solids the bases are reciprocally
518proportional to the heights; and those parallelepipedal solids in
519which the bases are reciprocally proportional to the heights are
520equal.
521\end{claim}
522\begin{evidence}[Proof of XI.34]
523\label{ev:XI.34}
5243D analogue of VI.14: by XI.32 the ratio of volumes is the
525compounded ratio of bases and heights; equal volumes force the
526compounded ratio to be unity, i.e.\ reciprocal proportion.
527\dependson{XI.34}{VI.14}
528\dependson{XI.34}{XI.32}
529\dependson{XI.34}{XI.33}
530\end{evidence}
531
532\begin{claim}[Proposition XI.35: Planes equally inclined to a base have equal perpendiculars]
533\label{prop:XI.35}
534If there be two equal plane angles, and on their vertices there be
535set up elevated straight lines containing equal angles with the
536original straight lines respectively, if on the elevated straight
537lines points be taken at random and perpendiculars be drawn from them
538to the planes in which the original angles are, and if from the
539points so arising in the planes straight lines be joined to the
540vertices of the original angles, they will contain, with the elevated
541straight lines, equal angles.
542\end{claim}
543\begin{evidence}[Proof of XI.35]
544\label{ev:XI.35}
545By I.4 (SAS) applied to the right triangles in each plane: equal
546oblique segments and equal perpendiculars produce equal angles at
547the foot.
548\dependson{XI.35}{I.4}
549\dependson{XI.35}{XI.10}
550\dependson{XI.35}{XI.11}
551\end{evidence}
552
553\begin{claim}[Proposition XI.36: Parallelepiped on three proportionals is equal to a cube whose side is the mean]
554\label{prop:XI.36}
555If three straight lines be proportional, the parallelepipedal solid
556formed out of the three is equal to the parallelepipedal solid on the
557mean which is equilateral, but equiangular with the aforesaid solid.
558\end{claim}
559\begin{evidence}[Proof of XI.36]
560\label{ev:XI.36}
561For $a : b = b : c$ (with $b$ the mean), $abc = b^3$ and the
562parallelepipeds on $\{a, b, c\}$ versus $\{b, b, b\}$ have equal
563volumes by XI.34 / XI.33.
564\dependson{XI.36}{VI.17}
565\dependson{XI.36}{XI.32}
566\dependson{XI.36}{XI.34}
567\end{evidence}
568
569\begin{claim}[Proposition XI.37: Similar parallelepipeds in proportion]
570\label{prop:XI.37}
571If four straight lines be proportional, the similar and similarly
572described parallelepipedal solids upon them will also be
573proportional; and if the similar and similarly described
574parallelepipedal solids upon them be proportional, the straight lines
575will themselves also be proportional.
576\end{claim}
577\begin{evidence}[Proof of XI.37]
578\label{ev:XI.37}
5793D analogue of VI.22: similar parallelepipeds are in the triplicate
580ratio of edges; equality of triplicate ratios is equivalent to
581equality of edge ratios.
582\dependson{XI.37}{VI.22}
583\dependson{XI.37}{XI.33}
584\end{evidence}
585
586\begin{claim}[Proposition XI.38: Joining diagonals of opposite faces in a cube]
587\label{prop:XI.38}
588If the sides of the opposite planes of a cube be bisected, and planes
589be carried through the points of section, the common section of the
590planes and the diameter of the cube bisect one another.
591\end{claim}
592\begin{evidence}[Proof of XI.38]
593\label{ev:XI.38}
594By symmetry of the cube under the half-turns about the centre; the
595diagonal and the median plane both pass through the centre.
596\dependson{XI.38}{I.10}
597\dependson{XI.38}{XI.3}
598\dependson{XI.38}{XI.24}
599\end{evidence}
600
601\begin{claim}[Proposition XI.39: Prisms on equal triangular bases with same height are equal]
602\label{prop:XI.39}
603If there be two prisms of equal height, and one have a parallelogram
604as base and the other a triangle, and if the parallelogram be double
605of the triangle, the prisms will be equal.
606\end{claim}
607\begin{evidence}[Proof of XI.39]
608\label{ev:XI.39}
609A triangular prism is half a parallelogram prism on the same height
610(by I.34 applied to the cross-sections); the area-doubling condition
611makes the two prisms have equal volume.
612\dependson{XI.39}{I.34}
613\dependson{XI.39}{XI.28}
614\dependson{XI.39}{XI.32}
615\end{evidence}
616