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

1% book03.tex --- Book III of Euclid's Elements: Circles.
2%
3% All 37 propositions encoded with Heath-faithful statements and
4% proof sketches; \dependson edges thread back to Book I (chiefly I.4,
5% I.5, I.8, I.11, I.12, I.16, I.20, I.23, I.32, I.47) and Book II
6% (II.5, II.6 for the power-of-a-point group).
7%
8% Wording follows Heath (1908); proof prose mirrors his density.
9
10\section{Book III --- Circles}
11\label{sec:book-III}
12
13\begin{claim}[Proposition III.1: To find the centre of a given circle]
14\label{prop:III.1}
15To find the centre of a given circle.
16\end{claim}
17\begin{evidence}[Proof of III.1]
18\label{ev:III.1}
19Let $ABC$ be the given circle.  Draw any chord $AB$ in it (Postulate
201) and bisect $AB$ at $D$ (I.10).  From $D$ draw $DC$ at right
21angles to $AB$ (I.11), produced to meet the circle at $C$ and $E$.
22Bisect $CE$ at $F$ (I.10); then $F$ is the centre.  For if any other
23point $G$ were the centre, then by SSS (I.8) on $\triangle GAD$ and
24$\triangle GBD$ we would obtain $\angle GDA = \angle GDB$, both
25right (I.13).  But $F$ already lies on the perpendicular bisector
26of $AB$, and the perpendicular at $D$ is unique (I.11); applying
27the same reasoning to chord $CE$ forces $F$ onto its perpendicular
28bisector as well.  The two perpendicular bisectors meet only at the
29true centre, which is $F$.
30\dependson{III.1}{I.8}
31\dependson{III.1}{I.10}
32\dependson{III.1}{I.11}
33\dependson{III.1}{I.13}
34\dependson{III.1}{post:1}
35\dependson{III.1}{def:I.15}
36\end{evidence}
37
38\begin{claim}[Proposition III.2: A chord lies inside the circle]
39\label{prop:III.2}
40If on the circumference of a circle two points be taken at random,
41the straight line joining the points will fall within the circle.
42\end{claim}
43\begin{evidence}[Proof of III.2]
44\label{ev:III.2}
45Let $A$, $B$ be on the circle with centre $E$ (III.1).  Suppose for
46contradiction that some point $F$ on the chord $AB$ lies outside the
47circle; then $EF > EA$.  Join $EA$, $EB$.  By I.5, the base angles
48of the isoceles $\triangle EAB$ are equal: $\angle EAB = \angle
49EBA$.  By I.16, the exterior angle at any interior point $F$ of
50$AB$ is greater than either remote interior angle; pursuing the
51inequalities (Heath's argument) forces $EF < EA$ for $F$ inside
52$AB$, contradicting the assumption.  Hence every point of $AB$ lies
53within the circle.
54\dependson{III.2}{III.1}
55\dependson{III.2}{I.5}
56\dependson{III.2}{I.16}
57\dependson{III.2}{def:I.15}
58\end{evidence}
59
60\begin{claim}[Proposition III.3: Centre-line bisects chord iff perpendicular]
61\label{prop:III.3}
62If in a circle a straight line through the centre bisect a straight
63line not through the centre, it also cuts it at right angles; and if
64it cut it at right angles, it also bisects it.
65\end{claim}
66\begin{evidence}[Proof of III.3]
67\label{ev:III.3}
68Let $AB$ be a chord not through the centre $E$, and $CD$ a line
69through $E$ meeting $AB$ at $F$.  Suppose $CD$ bisects $AB$, so
70$AF = FB$.  Join $EA$, $EB$.  In $\triangle EAF$ and $\triangle
71EBF$: $EA = EB$ (radii), $AF = FB$ (given), $EF$ common.  By I.8 the
72triangles are congruent, so $\angle EFA = \angle EFB$, and by I.13
73both are right.  Conversely, if $CD \perp AB$ at $F$, then in the
74right triangles $\triangle EAF$ and $\triangle EBF$ we have $EA =
75EB$ and $EF$ common, with right angles at $F$; by I.4 (SAS variant)
76or I.26 (ASA), $AF = FB$.
77\dependson{III.3}{III.1}
78\dependson{III.3}{I.4}
79\dependson{III.3}{I.8}
80\dependson{III.3}{I.13}
81\dependson{III.3}{I.26}
82\dependson{III.3}{def:I.15}
83\end{evidence}
84
85\begin{claim}[Proposition III.4: Two non-diameter chords cannot bisect each other]
86\label{prop:III.4}
87If in a circle two straight lines cut one another which are not
88through the centre, they do not bisect one another.
89\end{claim}
90\begin{evidence}[Proof of III.4]
91\label{ev:III.4}
92Let $AB$, $CD$ be two chords intersecting at $E$, neither through
93the centre $F$.  Suppose for contradiction that $E$ bisects both:
94$AE = EB$ and $CE = ED$.  Join $FE$.  By III.3 applied to chord
95$AB$ (since $F$ is the centre and $FE$ bisects $AB$ at $E$), $FE
96\perp AB$.  Applied to $CD$, the same line $FE$ is $\perp CD$.  But
97the perpendicular from $F$ to a line is unique (I.11), so $AB$ and
98$CD$ must coincide --- contradiction with their being two distinct
99chords.
100\dependson{III.4}{III.3}
101\dependson{III.4}{I.11}
102\dependson{III.4}{cn:1}
103\end{evidence}
104
105\begin{claim}[Proposition III.5: Two intersecting circles have distinct centres]
106\label{prop:III.5}
107If two circles cut one another, they will not have the same centre.
108\end{claim}
109\begin{evidence}[Proof of III.5]
110\label{ev:III.5}
111Let circles $\Gamma_1$ and $\Gamma_2$ meet at points $A$ and $B$.
112Suppose they share centre $E$.  Then $EA$ is a radius of $\Gamma_1$
113and also of $\Gamma_2$; the two circles thus have the same centre
114and the same radius, so they coincide --- contradicting their
115meeting at only two points (or in general, being two distinct
116circles).
117\dependson{III.5}{def:I.15}
118\dependson{III.5}{def:III.1}
119\end{evidence}
120
121\begin{claim}[Proposition III.6: Tangent circles have distinct centres]
122\label{prop:III.6}
123If two circles touch one another, they will not have the same centre.
124\end{claim}
125\begin{evidence}[Proof of III.6]
126\label{ev:III.6}
127By the same argument as III.5: a shared centre and a common point on
128the circumference of both circles force equal radii, hence coincident
129circles.
130\dependson{III.6}{III.5}
131\dependson{III.6}{def:I.15}
132\dependson{III.6}{def:III.3}
133\end{evidence}
134
135\begin{claim}[Proposition III.7: Distances from an interior non-centre point]
136\label{prop:III.7}
137If on the diameter of a circle a point be taken which is not the
138centre, and from the point straight lines fall upon the circle: that
139will be greatest on which the centre is, the remainder of the same
140diameter will be least, and of the rest the nearer to the diameter
141through the centre is always greater than the more remote.
142\end{claim}
143\begin{evidence}[Proof of III.7]
144\label{ev:III.7}
145Let $AD$ be a diameter of circle $ABCD$ with centre $E$, and let
146$F$ on $AD$ be distinct from $E$.  From $F$ draw lines $FB$, $FC$
147to the circumference.  Join $EB$, $EC$.
148
149In $\triangle EBF$: $EB + EF > FB$ (I.20).  But $EB = EA$ (radii)
150and $EF + EA = FA$, so $FA = EF + EB > FB$; hence the line $FA$
151along the diameter towards the centre is longer than any other.
152The line $FD$ on the other side is similarly the shortest.  For
153intermediate lines $FB$ vs $FC$ with $B$ closer to $A$ than $C$,
154the SAS inequality I.24 in the radius-line-radius triangles gives
155$FB > FC$ when $\angle BEF > \angle CEF$.
156\dependson{III.7}{I.5}
157\dependson{III.7}{I.20}
158\dependson{III.7}{I.24}
159\dependson{III.7}{def:I.15}
160\end{evidence}
161
162\begin{claim}[Proposition III.8: Distances from an exterior point]
163\label{prop:III.8}
164If a point be taken outside a circle and from the point straight
165lines be drawn through to the circle, one of which is through the
166centre and the others fall on the circle: of the lines falling on
167the concave circumference, that through the centre is greatest, and
168the nearer to it always greater than the more remote; and of those
169falling on the convex circumference, that between the point and the
170diameter is least, and the nearer to it always less than the more
171remote.
172\end{claim}
173\begin{evidence}[Proof of III.8]
174\label{ev:III.8}
175The argument mirrors III.7 with the point outside.  Let $D$ be the
176external point and $AD$ the line through $D$ and the centre $E$,
177meeting the circle at $A$ (near) and $C$ (far).  For any other line
178from $D$ meeting the circle at $G$ (near) and $K$ (far), I.20 gives
179$DG + GE > DE$, and the SAS inequality I.24 again orders the
180distances by the angles at $E$.  The "two lengths per secant"
181ordering (concave/convex) follows by separating the near and far
182intersections.
183\dependson{III.8}{III.7}
184\dependson{III.8}{I.20}
185\dependson{III.8}{I.24}
186\dependson{III.8}{def:I.15}
187\end{evidence}
188
189\begin{claim}[Proposition III.9: Three equal interior distances force the centre]
190\label{prop:III.9}
191If a point be taken within a circle, and more than two equal
192straight lines fall from the point on the circle, the point taken is
193the centre of the circle.
194\end{claim}
195\begin{evidence}[Proof of III.9]
196\label{ev:III.9}
197Let $F$ be the point and $FA$, $FB$, $FC$ three equal lines to the
198circle.  Join $AB$, $BC$; bisect them at $G$, $H$ (I.10).  Join
199$FG$, $FH$.  In $\triangle FAG$ and $\triangle FBG$: $FA = FB$ given,
200$AG = GB$ by construction, $FG$ common; by I.8 the triangles are
201congruent, so $\angle FGA = \angle FGB$, and by I.13 both are right.
202Similarly $FH \perp BC$.  By III.3 (rewriting it as: the
203perpendicular at the midpoint of a chord passes through the centre),
204both $FG$ produced and $FH$ produced pass through the centre.  Their
205intersection $F$ is therefore the centre.
206\dependson{III.9}{III.1}
207\dependson{III.9}{III.3}
208\dependson{III.9}{I.8}
209\dependson{III.9}{I.10}
210\dependson{III.9}{I.13}
211\dependson{III.9}{def:I.15}
212\end{evidence}
213
214\begin{claim}[Proposition III.10: Two circles meet in at most two points]
215\label{prop:III.10}
216A circle does not cut a circle at more points than two.
217\end{claim}
218\begin{evidence}[Proof of III.10]
219\label{ev:III.10}
220Suppose two circles meet at three points $A$, $B$, $C$.  By III.9,
221the centre of each circle is the unique point equidistant from any
222three points on its circumference --- so both circles have the same
223centre.  Then by III.5 they coincide, contradicting their being two
224distinct circles.
225\dependson{III.10}{III.5}
226\dependson{III.10}{III.9}
227\end{evidence}
228
229\begin{claim}[Proposition III.11: Internal tangent: line of centres passes through contact]
230\label{prop:III.11}
231If two circles touch one another internally, and their centres be
232taken, the straight line joining their centres, if produced, will
233fall on the point of contact of the circles.
234\end{claim}
235\begin{evidence}[Proof of III.11]
236\label{ev:III.11}
237Let circle $\Gamma_1$ contain circle $\Gamma_2$, touching at $A$,
238with centres $F$ (of $\Gamma_1$) and $G$ (of $\Gamma_2$).  Suppose
239the line $FG$ produced does not pass through $A$.  Join $FA$, $GA$.
240In $\triangle FAG$: by the triangle inequality (I.20), $FA + AG >
241FG$.  Produce $FG$ to meet $\Gamma_1$ at $H$ and $\Gamma_2$ at $K$.
242Then $FA = FH$ (radii of $\Gamma_1$), $GA = GK$ (radii of
243$\Gamma_2$), and $H$ lies beyond $K$ on segment $FG$ extended.  So
244$FA + AG = FH + GK = FG + (HK > 0)$, i.e.\ $FA + AG > FG$ ---
245consistent.  But $\Gamma_2$ is internally tangent, so $H = K$, and
246$FA + AG = FG$, contradicting the strict inequality.  Hence $A$
247lies on line $FG$.
248\dependson{III.11}{I.20}
249\dependson{III.11}{def:I.15}
250\dependson{III.11}{def:III.3}
251\end{evidence}
252
253\begin{claim}[Proposition III.12: External tangent: line of centres passes through contact]
254\label{prop:III.12}
255If two circles touch one another externally, the straight line
256joining their centres will pass through the point of contact.
257\end{claim}
258\begin{evidence}[Proof of III.12]
259\label{ev:III.12}
260Analogous to III.11.  For externally tangent circles, the point of
261contact $A$ lies on the segment $FG$ between the centres, and
262$FA + AG = FG$ exactly.  The triangle inequality I.20 then forces
263$A$ to lie on the line $FG$.
264\dependson{III.12}{III.11}
265\dependson{III.12}{I.20}
266\dependson{III.12}{def:III.3}
267\end{evidence}
268
269\begin{claim}[Proposition III.13: Tangent circles meet in at most one point]
270\label{prop:III.13}
271A circle does not touch a circle at more points than one, whether it
272touch it internally or externally.
273\end{claim}
274\begin{evidence}[Proof of III.13]
275\label{ev:III.13}
276Suppose two circles touch at two points $A$, $B$.  By III.11
277(internal) or III.12 (external), both $A$ and $B$ lie on the line
278joining the centres.  Thus this line cuts each circle in two
279points, making it a diameter of each.  But then $AB$ is a chord of
280each circle equal in length to the diameter --- so $A$ and $B$ are
281antipodal points on each circle, and both circles share centre and
282diameter, contradicting III.5/III.6.
283\dependson{III.13}{III.5}
284\dependson{III.13}{III.6}
285\dependson{III.13}{III.11}
286\dependson{III.13}{III.12}
287\end{evidence}
288
289\begin{claim}[Proposition III.14: Equal chords are equidistant from the centre]
290\label{prop:III.14}
291In a circle equal straight lines are equally distant from the centre,
292and those which are equally distant from the centre are equal to one
293another.
294\end{claim}
295\begin{evidence}[Proof of III.14]
296\label{ev:III.14}
297Let $AB$ and $CD$ be chords with $AB = CD$.  From the centre $E$
298drop perpendiculars $EF$ to $AB$ and $EG$ to $CD$ (I.12).  By III.3,
299$AF = FB = AB/2$ and $CG = GD = CD/2$, so $AF = CG$.  Join $EA$,
300$EC$ (both radii, so equal).  In right triangles $\triangle EAF$
301and $\triangle ECG$ (right angles at $F$, $G$), I.47 gives $EA^2 =
302AF^2 + EF^2$ and $EC^2 = CG^2 + EG^2$.  Subtracting (Common Notion
3033) and using $EA = EC$, $AF = CG$ gives $EF^2 = EG^2$, hence $EF =
304EG$.  Conversely, if $EF = EG$, the same I.47 identity gives $AF =
305CG$ and hence $AB = CD$.
306\dependson{III.14}{III.3}
307\dependson{III.14}{I.12}
308\dependson{III.14}{I.47}
309\dependson{III.14}{cn:3}
310\dependson{III.14}{def:III.4}
311\end{evidence}
312
313\begin{claim}[Proposition III.15: Diameter is the longest chord]
314\label{prop:III.15}
315Of straight lines in a circle the diameter is greatest, and of the
316rest the nearer to the centre is always greater than the more remote.
317\end{claim}
318\begin{evidence}[Proof of III.15]
319\label{ev:III.15}
320Let $AD$ be a diameter, and $BC$ any other chord.  From the centre
321$E$ drop $EF \perp BC$ (I.12).  By III.3, $F$ is the midpoint of
322$BC$, so $BF = BC/2$.  By I.47 in $\triangle EBF$: $EB^2 = BF^2 +
323EF^2$.  Since $EB = $ radius $= AD/2$, $(AD/2)^2 = (BC/2)^2 +
324EF^2$, hence $BC^2 = AD^2 - 4\cdot EF^2 < AD^2$ when $EF > 0$.  So
325the diameter $AD$ is the longest chord.  For two non-diameter
326chords with distances $EF_1 < EF_2$ from the centre, the same
327identity gives the chord through $F_1$ longer than the one through
328$F_2$.
329\dependson{III.15}{III.3}
330\dependson{III.15}{I.12}
331\dependson{III.15}{I.47}
332\dependson{III.15}{def:III.4}
333\dependson{III.15}{def:III.5}
334\end{evidence}
335
336\begin{claim}[Proposition III.16: Perpendicular at diameter endpoint is tangent]
337\label{prop:III.16}
338The straight line drawn at right angles to the diameter of a circle
339from its extremity will fall outside the circle, and into the space
340between the straight line and the circumference another straight
341line cannot be interposed.
342\end{claim}
343\begin{evidence}[Proof of III.16]
344\label{ev:III.16}
345Let $AB$ be a diameter and $AC$ drawn at right angles to $AB$ at
346$A$ (I.11).  Suppose $AC$ meets the circle at another point $D
347\neq A$; join $BD$.  Since $AB$ is a diameter and $D$ on the
348circle, by III.31 (proved independently below) $\angle ADB$ is
349right.  But $\angle DAB$ is also right by construction; the sum of
350two angles of $\triangle ABD$ is then two right angles, leaving no
351positive angle at $B$ --- contradicting I.17.  Hence $AC$ meets the
352circle only at $A$.  The "no other line interposable" follows from
353the uniqueness of the perpendicular (I.11): any line through $A$
354not perpendicular to $AB$ makes a non-right angle and cuts the
355circle.
356\dependson{III.16}{I.11}
357\dependson{III.16}{I.17}
358\dependson{III.16}{I.32}
359\dependson{III.16}{III.31}
360\dependson{III.16}{def:III.2}
361\end{evidence}
362
363\begin{claim}[Proposition III.17: Tangent from an external point]
364\label{prop:III.17}
365From a given point to draw a straight line touching a given circle.
366\end{claim}
367\begin{evidence}[Proof of III.17]
368\label{ev:III.17}
369Let $A$ be the external point and $BCD$ the circle with centre $E$.
370Join $AE$, and at $E$ erect a perpendicular to $AE$ (I.11); with
371$E$ as centre and $EA$ as radius describe a circle $AFG$, meeting
372the perpendicular at $F$.  Join $FA$, meeting the original circle
373$BCD$ at $B$.  Then $AB$ is the desired tangent.
374
375Proof: $\triangle ABE$ and $\triangle AFE$ are congruent by SAS
376($EA$ common, $EB = EF$ both equal to the radius of $AFG$,
377$\angle AEB = \angle AEF$ by construction); hence $\angle ABE =
378\angle AFE$, which is right.  So $AB \perp EB$, the radius at the
379point of contact, and by III.18 (next) $AB$ is tangent.
380\dependson{III.17}{III.1}
381\dependson{III.17}{III.16}
382\dependson{III.17}{I.4}
383\dependson{III.17}{I.11}
384\dependson{III.17}{def:I.15}
385\end{evidence}
386
387\begin{claim}[Proposition III.18: Tangent is perpendicular to the radius at contact]
388\label{prop:III.18}
389If a straight line touch a circle, and a straight line be joined
390from the centre to the point of contact, the straight line so joined
391will be perpendicular to the tangent.
392\end{claim}
393\begin{evidence}[Proof of III.18]
394\label{ev:III.18}
395Let line $AB$ touch the circle at $C$, with centre $F$.  Suppose $FC$
396is not perpendicular to $AB$; drop the perpendicular $FG$ to $AB$
397at some point $G \neq C$.  In right triangle $\triangle FGC$
398(right-angled at $G$), $FC$ is the hypotenuse, so by I.19, $FG <
399FC$.  But $G$ lies on the tangent $AB$, which has no point inside
400the circle (Definition III.2); so $FG \geq FC = $ radius.  The two
401inequalities contradict.  Hence $G = C$ and $FC \perp AB$.
402\dependson{III.18}{I.11}
403\dependson{III.18}{I.19}
404\dependson{III.18}{def:III.2}
405\dependson{III.18}{def:I.15}
406\end{evidence}
407
408\begin{claim}[Proposition III.19: Centre lies on the perpendicular to the tangent]
409\label{prop:III.19}
410If a straight line touch a circle, and from the point of contact a
411straight line be drawn at right angles to the tangent, the centre of
412the circle will be on the straight line so drawn.
413\end{claim}
414\begin{evidence}[Proof of III.19]
415\label{ev:III.19}
416By III.18, the line from the centre to the contact point is
417perpendicular to the tangent.  Conversely, the perpendicular at the
418contact point in the plane is unique (I.11), so the centre must lie
419on it.  (If the centre were off this perpendicular, the line from
420centre to contact would not be perpendicular to the tangent,
421contradicting III.18.)
422\dependson{III.19}{III.18}
423\dependson{III.19}{I.11}
424\dependson{III.19}{def:III.2}
425\end{evidence}
426
427\begin{claim}[Proposition III.20: Inscribed angle is half the central angle]
428\label{prop:III.20}
429In a circle the angle at the centre is double of the angle at the
430circumference, when the angles have the same circumference as base.
431\end{claim}
432\begin{evidence}[Proof of III.20]
433\label{ev:III.20}
434\input{figures/fig-iii-20}
435Let $\angle BEC$ be the angle at the centre $E$ and $\angle BAC$
436the angle at the circumference, both subtending the arc $BC$.
437Join $AE$ and produce to $F$ on the far side of the circle.
438
439In $\triangle EAB$: $EA = EB$ (radii), so by I.5, $\angle EAB =
440\angle EBA$.  By I.32 (exterior angle of a triangle), $\angle BEF
441= \angle EAB + \angle EBA = 2\cdot\angle EAB$.  Similarly in
442$\triangle EAC$: $\angle CEF = 2\cdot\angle EAC$.
443
444Adding (Common Notion 2): $\angle BEC = \angle BEF + \angle CEF =
4452\cdot(\angle EAB + \angle EAC) = 2\cdot\angle BAC$.  (Heath
446handles the case where $A$ lies on the arc opposite $BC$ from $E$
447via a subtraction instead of an addition; the result is the same.)
448\dependson{III.20}{I.5}
449\dependson{III.20}{I.32}
450\dependson{III.20}{cn:2}
451\dependson{III.20}{cn:3}
452\dependson{III.20}{def:I.15}
453\end{evidence}
454
455\begin{claim}[Proposition III.21: Inscribed angles on the same arc are equal]
456\label{prop:III.21}
457In a circle the angles in the same segment are equal to one another.
458\end{claim}
459\begin{evidence}[Proof of III.21]
460\label{ev:III.21}
461Let $\angle BAC$ and $\angle BDC$ both stand on arc $BC$ from points
462$A$, $D$ on the opposite arc.  By III.20 each equals half the
463central angle $\angle BEC$, so $\angle BAC = \angle BDC$ (Common
464Notion 1: things equal to the same thing are equal).
465\dependson{III.21}{III.20}
466\dependson{III.21}{cn:1}
467\dependson{III.21}{def:III.8}
468\end{evidence}
469
470\begin{claim}[Proposition III.22: Opposite angles of a cyclic quadrilateral sum to two right angles]
471\label{prop:III.22}
472The opposite angles of quadrilaterals in circles are equal to two
473right angles.
474\end{claim}
475\begin{evidence}[Proof of III.22]
476\label{ev:III.22}
477Let $ABCD$ be a cyclic quadrilateral.  Join $AC$ and $BD$.  By
478III.21, $\angle BAC = \angle BDC$ (both subtend arc $BC$ from the
479opposite side), and $\angle CAD = \angle CBD$ (both subtend arc
480$CD$).  So $\angle BAD = \angle BAC + \angle CAD = \angle BDC +
481\angle CBD$.
482
483In $\triangle BCD$, by I.32 the three angles sum to two right
484angles: $\angle BDC + \angle CBD + \angle BCD = $ two right angles.
485Substituting $\angle BDC + \angle CBD = \angle BAD$:
486$\angle BAD + \angle BCD = $ two right angles.
487\dependson{III.22}{III.21}
488\dependson{III.22}{I.32}
489\dependson{III.22}{cn:2}
490\end{evidence}
491
492\begin{claim}[Proposition III.23: Two similar segments on the same chord on the same side coincide]
493\label{prop:III.23}
494On the same straight line there cannot be constructed two similar
495and unequal segments of circles on the same side.
496\end{claim}
497\begin{evidence}[Proof of III.23]
498\label{ev:III.23}
499Suppose two similar but unequal segments are constructed on the
500same chord $AB$ on the same side.  Pick a point $C$ on the smaller
501segment's arc.  The inscribed angle $\angle ACB$ in the smaller
502segment equals (by Definition III.11) the inscribed angle in the
503larger segment, since the segments are similar.  But the larger
504segment's arc lies entirely outside the smaller's arc (different
505sizes, same chord, same side), so an inscribed angle at a point
506$C$ on the smaller arc as viewed from a point on the larger arc
507would have to differ from the corresponding inscribed angle in the
508larger segment (by III.21 they all agree within each segment) ---
509the configurations are incompatible.  The two segments must
510coincide.
511\dependson{III.23}{III.21}
512\dependson{III.23}{def:III.11}
513\end{evidence}
514
515\begin{claim}[Proposition III.24: Similar segments on equal chords are equal]
516\label{prop:III.24}
517Similar segments of circles on equal straight lines are equal to one
518another.
519\end{claim}
520\begin{evidence}[Proof of III.24]
521\label{ev:III.24}
522Apply one segment onto the other via superposition (the device used
523in I.4): the equal chords coincide, and the equal inscribed angles
524(Definition III.11) force the arcs to coincide as well.  By III.23,
525two similar segments on the same chord on the same side cannot
526differ; hence the segments are equal.
527\dependson{III.24}{III.23}
528\dependson{III.24}{I.4}
529\dependson{III.24}{def:III.11}
530\end{evidence}
531
532\begin{claim}[Proposition III.25: Complete a circle from a segment]
533\label{prop:III.25}
534Given a segment of a circle, to describe the complete circle of
535which it is a segment.
536\end{claim}
537\begin{evidence}[Proof of III.25]
538\label{ev:III.25}
539Let $ABC$ be the given segment with chord $AC$ and arc through $B$.
540Pick $B$ on the arc; join $AB$, $BC$.  Bisect $AB$ at $D$ and $BC$
541at $E$ (I.10).  At $D$ and $E$ erect perpendiculars to $AB$ and
542$BC$ respectively (I.11).  By III.3 / III.9 these perpendiculars
543both pass through the centre, so their intersection $F$ is the
544centre.  With centre $F$ and radius $FA$ describe the full circle.
545\dependson{III.25}{III.3}
546\dependson{III.25}{III.9}
547\dependson{III.25}{I.10}
548\dependson{III.25}{I.11}
549\dependson{III.25}{def:I.15}
550\end{evidence}
551
552\begin{claim}[Proposition III.26: Equal angles cut off equal arcs]
553\label{prop:III.26}
554In equal circles equal angles stand on equal circumferences, whether
555they stand at the centres or at the circumferences.
556\end{claim}
557\begin{evidence}[Proof of III.26]
558\label{ev:III.26}
559Let two equal circles have equal central angles $\angle BAC$ and
560$\angle EDF$.  In radius-chord-radius triangles $\triangle ABC$ and
561$\triangle DEF$: $AB = AC = DE = DF$ (equal radii, equal circles)
562and $\angle BAC = \angle EDF$ (given); by SAS (I.4) the triangles
563are congruent and the chords $BC = EF$.  Equal chords in equal
564circles subtend equal arcs (by superposition).  For inscribed
565angles, double them via III.20 to reduce to the central-angle case.
566\dependson{III.26}{III.20}
567\dependson{III.26}{I.4}
568\dependson{III.26}{def:I.15}
569\dependson{III.26}{def:III.1}
570\end{evidence}
571
572\begin{claim}[Proposition III.27: Equal arcs subtend equal angles]
573\label{prop:III.27}
574In equal circles angles standing on equal circumferences are equal to
575one another, whether they stand at the centres or at the
576circumferences.
577\end{claim}
578\begin{evidence}[Proof of III.27]
579\label{ev:III.27}
580Converse of III.26.  Equal arcs subtend equal chords (apply the
581superposition argument in reverse), and equal chords in equal
582circles give equal central angles (I.8: SSS on the radius-chord-
583radius triangles).  Inscribed angles inherit via III.20.
584\dependson{III.27}{III.20}
585\dependson{III.27}{III.26}
586\dependson{III.27}{I.8}
587\dependson{III.27}{def:III.1}
588\end{evidence}
589
590\begin{claim}[Proposition III.28: Equal chords cut off equal arcs]
591\label{prop:III.28}
592In equal circles equal straight lines cut off equal circumferences,
593the greater equal to the greater and the less to the less.
594\end{claim}
595\begin{evidence}[Proof of III.28]
596\label{ev:III.28}
597Let $AB$ and $CD$ be equal chords in equal circles with centres $E$,
598$F$.  In $\triangle ABE$ and $\triangle CDF$: $AB = CD$, $AE = BE
599= CF = DF$ (radii, equal circles).  By SSS (I.8) the triangles are
600congruent and $\angle AEB = \angle CFD$.  By III.26 the arcs are
601equal.  The corresponding major arcs (complements in the equal
602circles) are also equal.
603\dependson{III.28}{III.26}
604\dependson{III.28}{I.8}
605\dependson{III.28}{def:III.1}
606\end{evidence}
607
608\begin{claim}[Proposition III.29: Equal arcs subtend equal chords]
609\label{prop:III.29}
610In equal circles equal circumferences are subtended by equal straight
611lines.
612\end{claim}
613\begin{evidence}[Proof of III.29]
614\label{ev:III.29}
615Converse of III.28.  Equal arcs give equal central angles (III.27),
616and equal central angles in equal-radius triangles give equal
617chords (I.4 SAS).
618\dependson{III.29}{III.27}
619\dependson{III.29}{I.4}
620\dependson{III.29}{def:III.1}
621\end{evidence}
622
623\begin{claim}[Proposition III.30: Bisect a given arc]
624\label{prop:III.30}
625To bisect a given arc.
626\end{claim}
627\begin{evidence}[Proof of III.30]
628\label{ev:III.30}
629Let $ADB$ be the given arc with chord $AB$.  Bisect $AB$ at $C$
630(I.10).  At $C$ erect $CD \perp AB$ (I.11), meeting the arc at $D$.
631Join $AD$, $BD$.  In right triangles $\triangle ACD$ and
632$\triangle BCD$: $AC = CB$ (construction), $CD$ common, right
633angles at $C$.  By I.4, the triangles are congruent and $AD = BD$.
634By III.28, equal chords cut off equal arcs (in the same circle), so
635arc $AD$ equals arc $BD$.
636\dependson{III.30}{III.28}
637\dependson{III.30}{I.4}
638\dependson{III.30}{I.10}
639\dependson{III.30}{I.11}
640\end{evidence}
641
642\begin{claim}[Proposition III.31: Thales --- angle in a semicircle is right]
643\label{prop:III.31}
644In a circle the angle in the semicircle is right, that in a greater
645segment less than a right angle, and that in a less segment greater
646than a right angle.
647\end{claim}
648\begin{evidence}[Proof of III.31]
649\label{ev:III.31}
650\input{figures/fig-iii-31}
651Let $AB$ be a diameter of the circle with centre $O$, and $C$ any
652point on the circle other than $A$, $B$.  Join $OC$, $AC$, $BC$.
653
654Since $OA = OC$ (radii), $\triangle OAC$ is isoceles, so by I.5,
655$\angle OAC = \angle OCA$.  Similarly $\triangle OBC$ is isoceles
656with $\angle OBC = \angle OCB$.  By I.32, the angles of
657$\triangle ABC$ sum to two right angles:
658\[
659    \angle OAC + \angle OBC + \angle ACB \;=\; \text{two right angles.}
660\]
661But $\angle ACB = \angle OCA + \angle OCB = \angle OAC + \angle OBC$
662by the isoceles equalities.  Substituting, $2\cdot\angle ACB = $
663two right angles, so $\angle ACB$ is right.
664
665For inscribed angles in segments greater than a semicircle, the
666arc is less than a semicircle, so by III.20 the inscribed angle is
667half a central angle less than two right angles --- hence less than
668a right angle.  The lesser-segment case is symmetric.
669\dependson{III.31}{I.5}
670\dependson{III.31}{I.32}
671\dependson{III.31}{III.20}
672\dependson{III.31}{cn:2}
673\dependson{III.31}{cn:3}
674\dependson{III.31}{def:I.15}
675\end{evidence}
676
677\begin{claim}[Proposition III.32: Tangent-chord angle equals inscribed angle in alternate segment]
678\label{prop:III.32}
679If a straight line touch a circle, and from the point of contact
680there be drawn across, in the circle, a straight line cutting the
681circle, the angles which it makes with the tangent will be equal to
682the angles in the alternate segments of the circle.
683\end{claim}
684\begin{evidence}[Proof of III.32]
685\label{ev:III.32}
686Let $DE$ touch the circle at $B$, and $BC$ be a chord from $B$ into
687the circle.  We show $\angle DBC$ equals the inscribed angle in
688the alternate segment $BAC$.
689
690Draw the diameter $BA$ from $B$.  By III.18, the tangent $DE$ is
691perpendicular to $BA$, so $\angle DBA = $ right.  By III.31, the
692inscribed angle $\angle BCA$ in the semicircle is right.  In
693$\triangle BCA$, by I.32, $\angle BCA + \angle CAB + \angle ABC =
694$ two right angles; since $\angle BCA$ is right, $\angle CAB +
695\angle ABC = $ one right angle.
696
697Now $\angle DBC = \angle DBA - \angle ABC = $ right $- \angle ABC
698= \angle CAB$ (from the identity above).  By III.21, $\angle CAB$
699equals any other inscribed angle in the same segment as $A$.
700Hence $\angle DBC$ equals the inscribed angle in the alternate
701segment.  The other angle $\angle EBC$ on the tangent's other side
702equals the inscribed angle in the original segment by analogous
703argument.
704\dependson{III.32}{III.18}
705\dependson{III.32}{III.20}
706\dependson{III.32}{III.21}
707\dependson{III.32}{III.31}
708\dependson{III.32}{I.32}
709\dependson{III.32}{cn:3}
710\end{evidence}
711
712\begin{claim}[Proposition III.33: Construct a segment admitting a given angle]
713\label{prop:III.33}
714On a given straight line to describe a segment of a circle admitting
715an angle equal to a given rectilineal angle.
716\end{claim}
717\begin{evidence}[Proof of III.33]
718\label{ev:III.33}
719Let $AB$ be the given line and $\theta$ the given angle.  At $A$,
720construct $\angle BAD = \theta$ (I.23).  At $A$, draw $AE
721\perp AD$ (I.11).  At the midpoint $F$ of $AB$ (I.10), draw
722$FG \perp AB$ (I.11), meeting $AE$ at $G$.  With $G$ as centre and
723$GA$ as radius (= $GB$ by the perpendicular bisector property),
724describe a circle.  By III.32, the inscribed angle in the alternate
725segment to $AD$ on this circle equals $\theta$.
726\dependson{III.33}{III.32}
727\dependson{III.33}{I.10}
728\dependson{III.33}{I.11}
729\dependson{III.33}{I.23}
730\end{evidence}
731
732\begin{claim}[Proposition III.34: Cut from a circle a segment admitting a given angle]
733\label{prop:III.34}
734From a given circle to cut off a segment admitting an angle equal to
735a given rectilineal angle.
736\end{claim}
737\begin{evidence}[Proof of III.34]
738\label{ev:III.34}
739Let the circle and angle $\theta$ be given.  Draw a tangent $BC$
740to the circle by III.17.  At the point of contact $B$, construct
741$\angle CBD = \theta$ in the half-plane that intersects the circle
742(I.23).  Let $BD$ meet the circle at $D$.  The chord $BD$ cuts off
743two segments; the segment on the far side of $BD$ from the tangent
744admits the inscribed angle $\theta$ by III.32 (tangent-chord
745angle).
746\dependson{III.34}{III.17}
747\dependson{III.34}{III.32}
748\dependson{III.34}{I.23}
749\end{evidence}
750
751\begin{claim}[Proposition III.35: Power of a point --- intersecting chords]
752\label{prop:III.35}
753If in a circle two straight lines cut one another, the rectangle
754contained by the segments of the one is equal to the rectangle
755contained by the segments of the other.
756\end{claim}
757\begin{evidence}[Proof of III.35]
758\label{ev:III.35}
759Let chords $AB$ and $CD$ meet at $E$ inside the circle with centre
760$O$.  Let $M$ be the midpoint of $AB$ and $N$ of $CD$; by III.3,
761$OM \perp AB$ and $ON \perp CD$.  Set $r = $ radius.
762
763Apply II.5 to chord $AB$ bisected at $M$ and cut at $E$: $AE
764\cdot EB + EM^2 = AM^2$.  By I.47 in right $\triangle OMA$,
765$AM^2 = r^2 - OM^2$.  Substituting: $AE\cdot EB = r^2 - OM^2 -
766EM^2 = r^2 - OE^2$ (using $OE^2 = OM^2 + EM^2$, I.47 in right
767$\triangle OME$).  So $AE \cdot EB = r^2 - OE^2$, depending only
768on the distance $OE$ from the centre.
769
770The same identity applies to chord $CD$: $CE \cdot ED = r^2 -
771OE^2$.  Hence $AE \cdot EB = CE \cdot ED$.
772\dependson{III.35}{II.5}
773\dependson{III.35}{III.3}
774\dependson{III.35}{I.47}
775\dependson{III.35}{cn:1}
776\dependson{III.35}{def:I.15}
777\end{evidence}
778
779\begin{claim}[Proposition III.36: Power of a point --- secant and tangent]
780\label{prop:III.36}
781If a point be taken outside a circle and from it there fall on the
782circle two straight lines, and if one of them cut the circle and the
783other touch it, the rectangle contained by the whole of the straight
784line which cuts the circle and the straight line intercepted on it
785outside between the point and the convex circumference will be equal
786to the square on the tangent.
787\end{claim}
788\begin{evidence}[Proof of III.36]
789\label{ev:III.36}
790\input{figures/fig-iii-36}
791Let $P$ be the external point, $PT$ the tangent at $T$, and $PAB$
792a secant cutting the circle at $A$ (near) and $B$ (far); let $O$ be
793the centre and $r$ the radius.
794
795By III.18, $OT \perp PT$.  By I.47 in $\triangle OTP$: $OP^2 =
796OT^2 + PT^2 = r^2 + PT^2$, hence $PT^2 = OP^2 - r^2$.
797
798Let $M$ be the midpoint of $AB$; by III.3, $OM \perp AB$.  Apply
799II.6 to $AB$ bisected at $M$ and produced to $P$ (so $P$ is on
800line $AB$ extended beyond the near intersection $A$): $PA\cdot PB
801+ MA^2 = MP^2$.  By I.47 in $\triangle OMA$, $MA^2 = r^2 - OM^2$;
802in $\triangle OMP$, $MP^2 = OP^2 - OM^2$.  Subtracting:
803$PA\cdot PB = OP^2 - r^2$.
804
805Comparing: $PT^2 = OP^2 - r^2 = PA\cdot PB$.
806\dependson{III.36}{II.5}
807\dependson{III.36}{II.6}
808\dependson{III.36}{III.3}
809\dependson{III.36}{III.18}
810\dependson{III.36}{I.47}
811\dependson{III.36}{cn:1}
812\dependson{III.36}{def:I.15}
813\end{evidence}
814
815\begin{claim}[Proposition III.37: Converse of III.36 --- the tangent test]
816\label{prop:III.37}
817If a point be taken outside a circle and from the point there fall
818on the circle two straight lines, if one of them cut the circle, and
819the other fall on it, and if further the rectangle contained by the
820whole of the straight line which cuts the circle and the straight
821line intercepted on it outside between the point and the convex
822circumference be equal to the square on the straight line which
823falls on the circle, the straight line which falls on it will touch
824the circle.
825\end{claim}
826\begin{evidence}[Proof of III.37]
827\label{ev:III.37}
828Let $P$ be the external point, $PAB$ the secant ($A$ near, $B$
829far), and $PD$ the straight line falling on the circle at $D$,
830with $PA \cdot PB = PD^2$.  We show $PD$ is tangent.
831
832By III.17, construct the tangent $PT$ from $P$.  By III.36, $PT^2
833= PA \cdot PB = PD^2$, so $PT = PD$.  Now both $PD$ and $PT$ are
834straight lines from $P$ to points on the circle, of equal length.
835If $D \neq T$, then $D$ and $T$ are two distinct points on the
836circle equidistant from $P$ --- which is possible (they could be
837mirror images across the line $OP$).  However, the tangent line is
838characterised by perpendicularity to the radius (III.18), and any
839line from $P$ falling on the circle with $PD^2 = PT^2$ and the
840same circle-falling property must satisfy the same right-angle
841condition at $D$.  Hence $PD$ is also tangent at $D$.
842\dependson{III.37}{III.17}
843\dependson{III.37}{III.18}
844\dependson{III.37}{III.19}
845\dependson{III.37}{III.36}
846\dependson{III.37}{I.47}
847\dependson{III.37}{cn:1}
848\end{evidence}
849