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

1% book06.tex --- Book VI of Euclid's Elements: Similar Figures.
2%
3% All 33 propositions encoded. Book VI applies the Eudoxean proportion
4% machinery of Book V to plane figures: triangles, parallelograms, and
5% similar polygons. Highlights are VI.2 (the intercept theorem / "basic
6% proportionality theorem"), VI.4-VI.8 (similarity criteria), VI.13
7% (mean proportional construction), VI.30 (golden section by
8% application of areas), and VI.31 (generalised Pythagoras).
9%
10% Wording follows Heath (1908).
11
12\section{Book VI --- Similar Figures}
13\label{sec:book-VI}
14
15\begin{claim}[Proposition VI.1: Triangles and parallelograms with the same height]
16\label{prop:VI.1}
17Triangles and parallelograms which are under the same height are to
18one another as their bases.
19\end{claim}
20\begin{evidence}[Proof of VI.1]
21\label{ev:VI.1}
22Repeated application of I.38 generates any multiple of a triangle on
23the corresponding multiple of the base; the resulting equimultiples
24test (V.5) is exactly the statement of proportionality.  The
25parallelogram version follows because each parallelogram is double
26its diagonal triangle (I.34, I.41).
27\dependson{VI.1}{I.34}
28\dependson{VI.1}{I.38}
29\dependson{VI.1}{I.41}
30\dependson{VI.1}{def:V.5}
31\end{evidence}
32
33\begin{claim}[Proposition VI.2: Parallel to a side cuts proportionally]
34\label{prop:VI.2}
35If a straight line be drawn parallel to one of the sides of a
36triangle, it will cut the sides of the triangle proportionally; and
37if the sides of the triangle be cut proportionally, the line joining
38the points of section will be parallel to the remaining side of the
39triangle.
40\end{claim}
41\begin{evidence}[Proof of VI.2]
42\label{ev:VI.2}
43Drop a parallel $DE$ from a point $D$ on $AB$ to a point $E$ on $AC$,
44parallel to $BC$.  Triangles $\triangle DBC$ and $\triangle DCE$
45share the same base $DE$ and lie between the same parallels (with
46$BE$, $DC$ as transversals through I.29, I.37), so they are equal in
47area.  Applying VI.1 to $\triangle ADE$ versus the equal-area
48companion triangles gives $AD : DB = AE : EC$.  The converse runs the
49argument in reverse via I.39.
50\dependson{VI.2}{I.29}
51\dependson{VI.2}{I.37}
52\dependson{VI.2}{I.38}
53\dependson{VI.2}{I.39}
54\dependson{VI.2}{VI.1}
55\dependson{VI.2}{V.11}
56\end{evidence}
57
58\begin{claim}[Proposition VI.3: Angle bisector cuts opposite side proportionally]
59\label{prop:VI.3}
60If an angle of a triangle be bisected and the straight line cutting
61the angle cut the base also, the segments of the base will have the
62same ratio as the remaining sides of the triangle.
63\end{claim}
64\begin{evidence}[Proof of VI.3]
65\label{ev:VI.3}
66Through $C$ draw $CE$ parallel to the bisector $AD$ (I.31), meeting
67$BA$ produced at $E$.  By alternate angles (I.29) and the bisection
68hypothesis, $\angle ACE = \angle AEC$, so $AE = AC$ (I.6).  Applying
69VI.2 to $\triangle BCE$ with $AD \parallel CE$: $BD : DC = BA : AE =
70BA : AC$.
71\dependson{VI.3}{I.6}
72\dependson{VI.3}{I.29}
73\dependson{VI.3}{I.31}
74\dependson{VI.3}{VI.2}
75\end{evidence}
76
77\begin{claim}[Proposition VI.4: Equiangular triangles have proportional sides]
78\label{prop:VI.4}
79In equiangular triangles the sides about the equal angles are
80proportional, and those are corresponding sides which subtend the
81equal angles.
82\end{claim}
83\begin{evidence}[Proof of VI.4]
84\label{ev:VI.4}
85Lay the equiangular triangles side by side so that one pair of equal
86angles coincides; the remaining vertices and bases yield a
87parallelogram by I.28.  Apply VI.2 to the new figure to derive the
88proportionality of the sides.
89\dependson{VI.4}{I.28}
90\dependson{VI.4}{I.32}
91\dependson{VI.4}{VI.2}
92\end{evidence}
93
94\begin{claim}[Proposition VI.5: SSS-similarity criterion]
95\label{prop:VI.5}
96If two triangles have their sides proportional, the triangles will be
97equiangular and will have those angles equal which the corresponding
98sides subtend.
99\end{claim}
100\begin{evidence}[Proof of VI.5]
101\label{ev:VI.5}
102Construct on the second triangle's base a triangle equiangular with
103the first (I.23); by VI.4 its other sides are determined by the
104proportion, and by I.8 (SSS) it coincides with the second triangle.
105Therefore the second triangle is equiangular with the first.
106\dependson{VI.5}{I.8}
107\dependson{VI.5}{I.23}
108\dependson{VI.5}{VI.4}
109\end{evidence}
110
111\begin{claim}[Proposition VI.6: SAS-similarity criterion]
112\label{prop:VI.6}
113If two triangles have one angle equal to one angle and the sides
114about the equal angles proportional, the triangles will be
115equiangular and will have those angles equal which the corresponding
116sides subtend.
117\end{claim}
118\begin{evidence}[Proof of VI.6]
119\label{ev:VI.6}
120Same scheme as VI.5: extend one triangle so as to match the second
121on the equal-angle pair (I.23), apply VI.4 to deduce the missing
122side, then I.4 (SAS) for congruence of the auxiliary triangle with
123the second.
124\dependson{VI.6}{I.4}
125\dependson{VI.6}{I.23}
126\dependson{VI.6}{VI.4}
127\dependson{VI.6}{VI.5}
128\end{evidence}
129
130\begin{claim}[Proposition VI.7: Mixed SAS-similarity criterion]
131\label{prop:VI.7}
132If two triangles have one angle equal to one angle, the sides about
133other angles proportional, and the remaining angles either both less
134or both not less than a right angle, the triangles will be
135equiangular and will have those angles equal about which the sides
136are proportional.
137\end{claim}
138\begin{evidence}[Proof of VI.7]
139\label{ev:VI.7}
140Construct, at the vertex of the angle whose sides are proportional,
141an angle equal to the corresponding angle in the second triangle
142(I.23).  The resulting auxiliary triangle agrees with the first in
143two angles (and hence all three, by I.32) and shares a side with the
144second; the constraint on the remaining angle being acute or obtuse
145ensures the construction is non-ambiguous (essentially eliminates
146the SSA failure case).
147\dependson{VI.7}{I.23}
148\dependson{VI.7}{I.32}
149\dependson{VI.7}{VI.4}
150\dependson{VI.7}{VI.6}
151\end{evidence}
152
153\begin{claim}[Proposition VI.8: Right triangle altitude similarity]
154\label{prop:VI.8}
155If in a right-angled triangle a perpendicular be drawn from the right
156angle to the base, the triangles adjoining the perpendicular are
157similar both to the whole and to one another.
158\end{claim}
159\begin{evidence}[Proof of VI.8]
160\label{ev:VI.8}
161The two sub-triangles each share an angle with the original (the
162non-right angle at $B$ or $C$) and both have a right angle (at the
163foot of the altitude and at the apex), so they are equiangular with
164the original by I.32, hence similar by VI.4.  By transitivity (V.11)
165they are similar to each other.
166\dependson{VI.8}{I.32}
167\dependson{VI.8}{VI.4}
168\dependson{VI.8}{V.11}
169\end{evidence}
170
171\begin{claim}[Proposition VI.9: Cut off any part of a given segment]
172\label{prop:VI.9}
173From a given straight line to cut off a prescribed part.
174\end{claim}
175\begin{evidence}[Proof of VI.9]
176\label{ev:VI.9}
177Lay off a separate transversal containing the prescribed number of
178equal units (I.3 repeatedly).  Join the far ends and draw parallels
179to that join through each unit division (I.31).  By VI.2 these
180parallels mark off equal fractions on the given line.
181\dependson{VI.9}{I.3}
182\dependson{VI.9}{I.31}
183\dependson{VI.9}{VI.2}
184\end{evidence}
185
186\begin{claim}[Proposition VI.10: Divide a segment in a given ratio]
187\label{prop:VI.10}
188To cut a given uncut straight line similarly to a given cut straight
189line.
190\end{claim}
191\begin{evidence}[Proof of VI.10]
192\label{ev:VI.10}
193Lay the cut transversal alongside the given line meeting at one
194endpoint; join the far ends and draw parallels to that join through
195each cut point (I.31).  By VI.2 the parallels reproduce the same
196ratios on the given line.
197\dependson{VI.10}{I.31}
198\dependson{VI.10}{VI.2}
199\end{evidence}
200
201\begin{claim}[Proposition VI.11: Third proportional]
202\label{prop:VI.11}
203To two given straight lines to find a third proportional.
204\end{claim}
205\begin{evidence}[Proof of VI.11]
206\label{ev:VI.11}
207Place the two given segments on lines making an angle.  Mark off the
208second segment beyond the first; draw a parallel to the closing
209segment through the far endpoint (I.31).  By VI.2 the new mark-off
210is the required third proportional.
211\dependson{VI.11}{I.31}
212\dependson{VI.11}{VI.2}
213\end{evidence}
214
215\begin{claim}[Proposition VI.12: Fourth proportional]
216\label{prop:VI.12}
217To three given straight lines to find a fourth proportional.
218\end{claim}
219\begin{evidence}[Proof of VI.12]
220\label{ev:VI.12}
221Same construction as VI.11 but with three input segments: lay two on
222one transversal and one on the other, then drop a parallel from the
223last point.  VI.2 yields the fourth proportional.
224\dependson{VI.12}{I.31}
225\dependson{VI.12}{VI.2}
226\end{evidence}
227
228\begin{claim}[Proposition VI.13: Mean proportional]
229\label{prop:VI.13}
230To two given straight lines to find a mean proportional.
231\end{claim}
232\begin{evidence}[Proof of VI.13]
233\label{ev:VI.13}
234Lay the two segments end-to-end as $AB$, $BC$ on a single line; on
235$AC$ as diameter describe a semicircle (Postulate 3, III.31 implicit).
236Erect the perpendicular $BD$ at $B$ to the diameter; then $\triangle
237ABD$, $\triangle BDC$, and $\triangle ABC$ are similar by VI.8, so
238$AB : BD = BD : BC$, making $BD$ the required mean proportional.
239\dependson{VI.13}{I.11}
240\dependson{VI.13}{III.31}
241\dependson{VI.13}{VI.8}
242\dependson{VI.13}{post:3}
243\end{evidence}
244
245\begin{claim}[Proposition VI.14: Equal parallelograms have reciprocally proportional sides]
246\label{prop:VI.14}
247In equal and equiangular parallelograms the sides about the equal
248angles are reciprocally proportional; and equiangular parallelograms
249in which the sides about the equal angles are reciprocally
250proportional are equal.
251\end{claim}
252\begin{evidence}[Proof of VI.14]
253\label{ev:VI.14}
254Lay the two parallelograms so the equal angles coincide; their union
255forms a third parallelogram whose diagonal includes the original
256common-angle vertex.  By VI.1 the ratios of areas equal the ratios of
257adjacent sides; equality of the original areas forces the
258reciprocal-proportion relation.  Converse runs the same way.
259\dependson{VI.14}{VI.1}
260\dependson{VI.14}{V.11}
261\end{evidence}
262
263\begin{claim}[Proposition VI.15: Equal triangles with one common angle have reciprocally proportional sides]
264\label{prop:VI.15}
265In equal triangles which have one angle equal to one angle the sides
266about the equal angles are reciprocally proportional; and those
267triangles which have one angle equal to one angle, and in which the
268sides about the equal angles are reciprocally proportional, are equal.
269\end{claim}
270\begin{evidence}[Proof of VI.15]
271\label{ev:VI.15}
272Same scheme as VI.14 applied to triangles (each is half a
273parallelogram, so the areas-of-parallelograms result transfers).
274\dependson{VI.15}{I.41}
275\dependson{VI.15}{VI.14}
276\end{evidence}
277
278\begin{claim}[Proposition VI.16: Equal rectangles iff proportional sides]
279\label{prop:VI.16}
280If four straight lines be proportional, the rectangle contained by the
281extremes is equal to the rectangle contained by the means; and if the
282rectangle contained by the extremes be equal to the rectangle
283contained by the means, the four straight lines will be proportional.
284\end{claim}
285\begin{evidence}[Proof of VI.16]
286\label{ev:VI.16}
287If $a : b = c : d$, the rectangles $ad$ and $bc$ are equiangular
288(all right-angled) and have sides reciprocally proportional, so by
289VI.14 they are equal.  Conversely from $ad = bc$ and VI.14 (its
290converse) we recover the proportion.
291\dependson{VI.16}{VI.14}
292\end{evidence}
293
294\begin{claim}[Proposition VI.17: Mean proportional iff equal squares]
295\label{prop:VI.17}
296If three straight lines be proportional, the rectangle contained by
297the extremes is equal to the square on the mean; and if the rectangle
298contained by the extremes be equal to the square on the mean, the
299three straight lines will be proportional.
300\end{claim}
301\begin{evidence}[Proof of VI.17]
302\label{ev:VI.17}
303Specialise VI.16 to $b = c$.
304\dependson{VI.17}{VI.16}
305\end{evidence}
306
307\begin{claim}[Proposition VI.18: Construct a polygon similar to a given polygon]
308\label{prop:VI.18}
309On a given straight line to describe a rectilineal figure similar
310and similarly situated to a given rectilineal figure.
311\end{claim}
312\begin{evidence}[Proof of VI.18]
313\label{ev:VI.18}
314Triangulate the given polygon by joining a vertex to all others.  On
315the new base copy each angle (I.23) and use VI.4 to fix the side
316ratios; assemble the triangles into the similar polygon.
317\dependson{VI.18}{I.23}
318\dependson{VI.18}{VI.4}
319\end{evidence}
320
321\begin{claim}[Proposition VI.19: Similar triangles are as the squares on corresponding sides]
322\label{prop:VI.19}
323Similar triangles are to one another in the duplicate ratio of the
324corresponding sides.
325\end{claim}
326\begin{evidence}[Proof of VI.19]
327\label{ev:VI.19}
328Let $\triangle ABC \sim \triangle DEF$ with side ratio $k = BC / EF$.
329Construct $G$ on $BC$ so that $BG = EF \cdot k$ (i.e.\ a third
330proportional, VI.11).  By VI.1 the area ratio is $BG : EF = k$ along
331one dimension and the side ratio $k$ along the perpendicular, giving
332total area ratio $k^2$ in the sense of Definition V.9 (duplicate
333ratio).
334\dependson{VI.19}{VI.1}
335\dependson{VI.19}{VI.11}
336\dependson{VI.19}{def:V.9}
337\end{evidence}
338
339\begin{claim}[Proposition VI.20: Similar polygons are as the squares on corresponding sides]
340\label{prop:VI.20}
341Similar polygons are divided into similar triangles, equal in
342multitude and in the same ratio as the wholes; and the polygon has
343to the polygon a ratio duplicate of that which the corresponding side
344has to the corresponding side.
345\end{claim}
346\begin{evidence}[Proof of VI.20]
347\label{ev:VI.20}
348Decompose into similar triangles by joining one vertex to all others;
349apply VI.19 to each triangle and sum via V.12.
350\dependson{VI.20}{VI.18}
351\dependson{VI.20}{VI.19}
352\dependson{VI.20}{V.12}
353\end{evidence}
354
355\begin{claim}[Proposition VI.21: Figures similar to the same are similar]
356\label{prop:VI.21}
357Figures which are similar to the same rectilineal figure are also
358similar to one another.
359\end{claim}
360\begin{evidence}[Proof of VI.21]
361\label{ev:VI.21}
362Equiangularity transfers transitively, and the side ratios compose
363transitively by V.11.
364\dependson{VI.21}{V.11}
365\dependson{VI.21}{VI.4}
366\end{evidence}
367
368\begin{claim}[Proposition VI.22: Proportionality of figures on proportional sides]
369\label{prop:VI.22}
370If four straight lines be proportional, the rectilineal figures
371similar and similarly described upon them will also be proportional;
372and if the rectilineal figures similar and similarly described upon
373them be proportional, the straight lines will themselves also be
374proportional.
375\end{claim}
376\begin{evidence}[Proof of VI.22]
377\label{ev:VI.22}
378Apply VI.20: figures similar on proportional sides have areas in the
379duplicate ratio of the sides.  Equality of those duplicate ratios is
380equivalent to equality of the original side ratios.
381\dependson{VI.22}{VI.20}
382\dependson{VI.22}{def:V.9}
383\end{evidence}
384
385\begin{claim}[Proposition VI.23: Ratio of parallelograms is compound of side ratios]
386\label{prop:VI.23}
387Equiangular parallelograms have to one another the ratio compounded
388of the ratios of their sides.
389\end{claim}
390\begin{evidence}[Proof of VI.23]
391\label{ev:VI.23}
392Place the two parallelograms so the equal angles share a vertex; the
393resulting figure can be split by lines parallel to the sides into a
394rectangle whose dimensions are the four sides.  Two applications of
395VI.1 give the compounded ratio.
396\dependson{VI.23}{VI.1}
397\dependson{VI.23}{def:V.10}
398\end{evidence}
399
400\begin{claim}[Proposition VI.24: Parallelograms about the diagonal are similar to the whole]
401\label{prop:VI.24}
402In any parallelogram the parallelograms about the diameter are
403similar both to the whole and to one another.
404\end{claim}
405\begin{evidence}[Proof of VI.24]
406\label{ev:VI.24}
407Lines drawn parallel to the sides through a point on the diagonal
408make the inner parallelograms equiangular with the whole (I.29) and
409with corresponding sides cut in the same ratio (VI.2); hence similar
410(VI.4).
411\dependson{VI.24}{I.29}
412\dependson{VI.24}{VI.2}
413\dependson{VI.24}{VI.4}
414\end{evidence}
415
416\begin{claim}[Proposition VI.25: Construct a figure similar to one and equal to another]
417\label{prop:VI.25}
418To construct one and the same figure similar to a given rectilineal
419figure and equal to another given rectilineal figure.
420\end{claim}
421\begin{evidence}[Proof of VI.25]
422\label{ev:VI.25}
423Reduce both given figures to rectangles on a common base (I.44);
424the side opposite the common base measures each figure's area.  Take
425the mean proportional (VI.13) of those two opposite sides; build on
426that mean a figure similar to the first via VI.18.  By VI.20 the
427constructed figure has the required area.
428\dependson{VI.25}{I.44}
429\dependson{VI.25}{VI.13}
430\dependson{VI.25}{VI.18}
431\dependson{VI.25}{VI.20}
432\end{evidence}
433
434\begin{claim}[Proposition VI.26: Inner parallelogram about the diagonal must share a corner]
435\label{prop:VI.26}
436If from a parallelogram there be taken away a parallelogram similar
437and similarly situated to the whole and having a common angle with
438it, it is about the same diameter with the whole.
439\end{claim}
440\begin{evidence}[Proof of VI.26]
441\label{ev:VI.26}
442Reductio: if the inner parallelogram is not about the diagonal, then
443VI.24 forces a similar parallelogram on the actual diagonal that
444differs from the given one; matching corresponding sides via the
445similarity contradicts the assumed common-angle configuration.
446\dependson{VI.26}{VI.24}
447\dependson{VI.26}{V.9}
448\end{evidence}
449
450\begin{claim}[Proposition VI.27: Maximum-area parallelogram with deficiency]
451\label{prop:VI.27}
452Of all parallelograms applied to the same straight line and deficient
453by parallelogrammic figures similar and similarly situated to that
454described upon the half of the straight line, the greatest is that
455which is applied to the half and is similar to the deficient figure.
456\end{claim}
457\begin{evidence}[Proof of VI.27]
458\label{ev:VI.27}
459Comparison of areas via VI.20 and VI.24: the application to the half
460exhausts the bound, while any other application produces a smaller
461parallelogram by a square-on-the-deviation.
462\dependson{VI.27}{VI.20}
463\dependson{VI.27}{VI.24}
464\dependson{VI.27}{VI.26}
465\end{evidence}
466
467\begin{claim}[Proposition VI.28: Apply a parallelogram with deficiency]
468\label{prop:VI.28}
469To a given straight line to apply a parallelogram equal to a given
470rectilineal figure and deficient by a parallelogrammic figure similar
471to a given one; thus the given rectilineal figure must not be greater
472than the parallelogram described on the half of the straight line and
473similar to the defect.
474\end{claim}
475\begin{evidence}[Proof of VI.28]
476\label{ev:VI.28}
477Construct the application by VI.25 to match the given figure, then
478verify the bound via VI.27.  The construction effectively solves a
479quadratic equation in geometric form.
480\dependson{VI.28}{VI.25}
481\dependson{VI.28}{VI.27}
482\end{evidence}
483
484\begin{claim}[Proposition VI.29: Apply a parallelogram with excess]
485\label{prop:VI.29}
486To a given straight line to apply a parallelogram equal to a given
487rectilineal figure and exceeding by a parallelogrammic figure similar
488to a given one.
489\end{claim}
490\begin{evidence}[Proof of VI.29]
491\label{ev:VI.29}
492Dual to VI.28; solves the geometric quadratic with the opposite sign.
493\dependson{VI.29}{VI.25}
494\dependson{VI.29}{VI.28}
495\end{evidence}
496
497\begin{claim}[Proposition VI.30: Golden section by application of areas]
498\label{prop:VI.30}
499To cut a given finite straight line in extreme and mean ratio.
500\end{claim}
501\begin{evidence}[Proof of VI.30]
502\label{ev:VI.30}
503Apply to the given line a parallelogram equal to the square on it,
504exceeding by a square (VI.29).  The exceeding square's side $x$
505satisfies $x(a+x) = a^2$, equivalently $a : x = x : (a - x)$ in the
506language of II.11 -- the golden section.
507\dependson{VI.30}{II.11}
508\dependson{VI.30}{VI.17}
509\dependson{VI.30}{VI.29}
510\end{evidence}
511
512\begin{claim}[Proposition VI.31: Generalised Pythagoras]
513\label{prop:VI.31}
514In right-angled triangles the figure on the side subtending the right
515angle is equal to the similar and similarly described figures on the
516sides containing the right angle.
517\end{claim}
518\begin{evidence}[Proof of VI.31]
519\label{ev:VI.31}
520By VI.8 the altitude from the right angle cuts the hypotenuse into
521segments such that each leg is the mean proportional between the
522hypotenuse and the adjacent segment.  Applying VI.20 (areas of
523similar figures are in the duplicate ratio of corresponding sides)
524gives each leg-figure equal to its adjacent piece of the
525hypotenuse-figure.  Summing the two pieces by V.24 yields the
526hypotenuse-figure.
527\dependson{VI.31}{VI.8}
528\dependson{VI.31}{VI.19}
529\dependson{VI.31}{VI.20}
530\dependson{VI.31}{V.24}
531\end{evidence}
532
533\begin{claim}[Proposition VI.32: Similarly placed triangles share a vertex angle]
534\label{prop:VI.32}
535If two triangles having two sides proportional to two sides be placed
536together at one angle so that their corresponding sides are also
537parallel, the remaining sides of the triangles will be in a straight
538line.
539\end{claim}
540\begin{evidence}[Proof of VI.32]
541\label{ev:VI.32}
542Equal angles and proportional sides (VI.6) force the triangles to
543share the same straight-line continuation at the meeting vertex (I.14).
544\dependson{VI.32}{I.14}
545\dependson{VI.32}{I.29}
546\dependson{VI.32}{VI.6}
547\end{evidence}
548
549\begin{claim}[Proposition VI.33: Inscribed angles are as their arcs]
550\label{prop:VI.33}
551In equal circles angles have the same ratio as the circumferences on
552which they stand, whether they stand at the centres or at the
553circumferences.
554\end{claim}
555\begin{evidence}[Proof of VI.33]
556\label{ev:VI.33}
557Use III.27 (equal arcs subtend equal central angles in equal circles)
558to set up an equimultiples test: for any positive integers $m$, $n$,
559$m$ copies of one angle correspond to $m$ copies of its arc, and the
560order of $m \alpha$ versus $n \beta$ matches the order of $m \cdot
561\mathrm{arc}(\alpha)$ versus $n \cdot \mathrm{arc}(\beta)$.  This is
562exactly Definition V.5 for $\alpha : \beta = \mathrm{arc}(\alpha) :
563\mathrm{arc}(\beta)$.  The inscribed-angle case follows from III.20
564(inscribed angle is half the central).
565\dependson{VI.33}{III.20}
566\dependson{VI.33}{III.27}
567\dependson{VI.33}{def:V.5}
568\end{evidence}
569