Relationships Between Circles Inscribed in Triangles and Related Curvilinear Triangles
SSangaku Journal of Mathematics (SJM) © SJMISSN 2534-9562Volume 4 (2020), pp.9-27Received 19 November 2019. Published on-line 16 January 2020web: © The Author(s) This article is published with open access . Relationships Between Circles Inscribed inTriangles and Related Curvilinear Triangles
Stanley Rabinowitz
545 Elm St Unit 1, Milford, New Hampshire 03055, USAe-mail: [email protected]:
Abstract. If P is a point inside (cid:52) ABC , then the cevians through P extendedto the circumcircle of (cid:52) ABC create a figure containing a number of curvilineartriangles. Each curvilinear triangle is bounded by an arc of the circumcircle andtwo line segments lying along the sides or cevians of the original triangle. We givetheorems about the relationships between the radii of circles inscribed in varioussets of these curvilinear triangles.
Keywords. circles, cevians, curvilinear triangles, sangaku.
Mathematics Subject Classification (2010).
Introduction A curvilinear triangle is a geometric figure bounded by three curves. The curvesare typically line segments and arcs of circles, in which case there is a unique circletangent to each of the three boundary curves. This circle is called the incircle ofthe curvilinear triangle.Wasan geometers loved to find relationships between the radii of circles inscribedin curvilinear triangles. An example is shown in Figure 1 which comes from an1841 book of Mathematical Formulae written by Yamamoto [12]. It is also givenas problem 5.3.9 in [4]. In the figure, AH ⊥ BC and BA ⊥ AC . There arethree curvilinear triangles of interest in the figure. The first curvilinear triangleis bounded by BH , HA , and arc (cid:95) AB . The second curvilinear triangle is boundedby AH , HC , and arc (cid:95) CA . The third curvilinear triangle is bounded by CA , AB ,and arc (cid:95) BC . The radii of the circles inscribed in these curvilinear triangles are r , r , and r , respectively. Then the nice relationship that was found is r + r = r . This article is distributed under the terms of the Creative Commons Attribution Licensewhich permits any use, distribution, and reproduction in any medium, provided the originalauthor(s) and the source are credited. a r X i v : . [ m a t h . HO ] J a n Relationships Between Circles Inscribed in Triangles and Curvilinear Triangles
Figure 1. r + r = r In this paper, we will find some other nice relationships between the inradii ofcurvilinear triangles.If a curvilinear triangle is convex and bounded by two straight line segmentsand one circular arc, then we will call the resulting figure a skewed sector (seeFigure 2).
Figure 2. a skewed sector
Anatomy of a skewed sector. • The two straight line segments are called the sides of the skewed sector. • The point of intersection of the two sides is called the vertex of the skewedsector. • The angle between the two sides is called the vertex angle . • The circular arc is referred to as the arc of the skewed sector . • The circular measure of the arc of a skewed sector is called the arc angle . • The circle to which the arc belongs will be called the circle associated withthe skewed sector . • The triangle formed by the vertex of a skewed sector and the endpoints ofits arc will be referred to as the triangle associated with the skewed sector .This would be (cid:52)
AP B in Figure 2. • When naming a skewed sector, the vertex will always be the middle letter.Thus, the skewed sector in Figure 2 is named skewed sector
AP B . tanley Rabinowitz • When the vertex of a skewed sector lies inside the associated circle, if thesides of the skewed sector are extended back through the vertex, they willintercept an arc of the associated circle. This arc is called the opposite arc of the skewed sector. It is shown in red in Figure 3. • The vertex of a skewed sector and the opposite arc form another skewedsector called the opposite skewed sector . This is skewed sector A (cid:48) P B (cid:48) inFigure 3.
Figure 3. opposite arcA segment of a circle is the figure bounded by an arc of a circle and the chordjoining the endpoints of that arc. The height (or sagitta) of the segment is thedistance from the midpoint of the chord to the midpoint of the arc.If P is a point inside (cid:52) ABC , then the cevians through P extended to the cir-cumcircle of (cid:52) ABC create a figure containing a number of skewed sectors. Wewill find relationships between the radii of the circles inscribed in some of theseskewed sectors.
Notation. • If X and Y are points, then we use the notation XY to denote either theline segment joining X and Y or the length of that line segment, dependingon the context. • A cevian of a triangle is a line segment from a vertex to the opposite side. • We use the notation ∠ XY Z to denote either the angle between XY and Y Z or the measure of that angle, depending on the context. • The notation [XYZ] denotes the area of (cid:52)
XY Z . • The notation O ( r ) refers to the circle centered at point O with radius r .The circle may sometimes also be referred to as circle O . • If (cid:95) XY is an arc of a circle, then m ( (cid:95) XY ) denotes the circular measure ofthat arc. The arc extends counterclockwise along the circle from X to Y . • Typically, we use r for the inradius of a triangle and w for the inradius ofa skewed sector. Relationships Between Circles Inscribed in Triangles and Curvilinear Triangles Inradius Formula
Formulas for the radius of the circle inscribed in a skewed sector and in a triangleare known. Since these are not well-known, we review them here.
Theorem 2.1 (Inradius of skewed sector) . Let
AP B be a skewed sector and let C be the circle associated with arc (cid:95) AB . Suppose P lies inside C . Let R be the radiusof C , let w be the radius of the circle inscribed in the skewed sector, and let r bethe radius of the circle inscribed in (cid:52) AP B . Extend BP to meet the circle C at C and draw AC . Let α , β , γ , δ , and (cid:15) be the measures of five angles associated withthe skewed sector as shown in Figure 4. Then w = 4 R sin β γ δ (cid:15) (cid:16) cos α (cid:17) ,r = 4 R sin β γ (cid:15) (cid:15) α . Figure 4. angles associated with a skewed sector
Proof.
See [4, pp. 96–97] or [11, p. 26]. (cid:3)
An immediate consequence of this theorem is the following result.
Theorem 2.2.
Let
AP B be a skewed sector and suppose P lies inside the asso-ciated circle. Let w be the radius of the circle inscribed in the skewed sector, andlet r be the radius of the circle inscribed in (cid:52) AP B . Let α be the vertex angle ofthe skewed sector, let θ be the arc angle of the skewed sector, and let θ be thearc angle of the opposite skewed sector. See Figure 5. Then wr = cos( θ / α/
2) cos( θ / . We can also express w/r without using θ as follows. tanley Rabinowitz Figure 5.
Theorem 2.3.
Using the same notation as in Theorem 2.2, wr = 1 + tan( α/
2) tan( θ / . Proof.
See [11, pp. 26–27]. (cid:3) Relationship Between Incircles of Skewed Sectors andIncircles of Triangles
To prove a relationship between skewed sector inradii, Theorems 2.1, 2.2, or 2.3could be used to find the length of each radius. This is a brute force techniqueand better methods are available. One strategy for finding relationships betweenthe radii of circles inscribed in skewed sectors is to relate these circles to circlesinscribed in triangles, for which results are already known.Here are some theorems that relate circles in skewed sectors to circles in triangles.The following theorem appeared on a tablet in 1781. See [4, problem 4.0.3], [3,problem 2.2.8], [5], and [9].
Theorem 3.1 (Ajima’s Theorem) . Let AB be a chord of a circle and let C be apoint inside the circle, not on the chord. See Figure 6. Let W ( w ) be the incircleof skewed sector ACB (the red circle) and let O ( r ) be the incircle of (cid:52) ACB (theyellow circle). Then w = r + 2 d ( s − a )( s − b ) cs , where d is the height of the segment formed by AB , a = BC , b = AC , c = AB ,and s is the semiperimeter of (cid:52) ABC .Proof.
See [4, pp. 96-97]. A more detailed proof can be found in [10, pp. 40-49]. (cid:3)
Theorem 3.2.
Let D be any point on side BC of (cid:52) ABC . Cevian AD extendedmeets the circumcircle of (cid:52) ABC at D (cid:48) . Let W ( w ) be the incircle of skewed Relationships Between Circles Inscribed in Triangles and Curvilinear Triangles
Figure 6. w = r + 2 d ( s − a )( s − b ) / ( cs ) sector BDD (cid:48) and let W ( w ) be the incircle of skewed sector CDD (cid:48) . Let O ( r ) bethe incircle of (cid:52) ADB and let O ( r ) be the incircle of (cid:52) ADC (Figure 7). Then r + 1 w = 1 r + 1 w . Figure 7. /r + 1 /w = 1 /r + 1 /w Proof.
We give names to the various angles as shown in Figure 8.Note that ∠ CDA = α is supplementary to ∠ BDA = α , so cos( α /
2) = sin( α/ w = 4 R sin( β/
2) cos( γ/
2) sin( δ/
2) sin( (cid:15)/ ( α/ ,w = 4 R sin( β/
2) sin( γ/
2) sin( δ/
2) cos( (cid:15)/ ( α/ ,r = 4 R sin( β/
2) sin( γ/
2) sin( (cid:15)/
2) cos( (cid:15)/ α/ ,r = 4 R sin( δ/
2) sin( (cid:15)/
2) sin( γ/
2) cos( γ/ α/ . tanley Rabinowitz Figure 8. angle namesNow form the expression S = 1 w + 1 r − (cid:18) w + 1 r (cid:19) . Then substitute α = δ + (cid:15) and then (cid:15) = π − β − γ − δ . Simplifying the resultingexpression (using a computer algebra system), shows that S = 0. (cid:3) The result of Theorem 3.2 is so elegant that it is unlikely that it is true onlybecause the complicated trigonometric expression, S , in the proof just happensto simplify to 0. Open Question.
Is there a simple proof of Theorem 3.2 that does not involve alarge amount of trigonometric computation requiring computer simplification?
The following theorem appeared on a tablet in 1844 in the Aichi prefecture. See[3, problem 1.4.7] and [2, p. 22].
Theorem 3.3.
Chords AB and CD of a circle meet at E . Let W ( w ) be theincircle of skewed sector BED and let W ( w ) be the incircle of skewed sector AEC . Let O ( r ) be the incircle of (cid:52) BED and let O ( r ) be the incircle of (cid:52) AEC . See Figure 9. Then r + 1 w = 1 r + 1 w . Proof.
This proof comes from [11, pp. 26–27]. Let α be the vertex angle of skewedsector BED . Let θ be its arc angle an let θ be the arc angle of the oppositeskewed sector AEC . By Theorem 2.3, we have w r = 1 + tan( α/
2) tan( θ / w r − α/
2) tan( θ / Relationships Between Circles Inscribed in Triangles and Curvilinear Triangles
Figure 9. /r + 1 /w = 1 /r + 1 /w or 1 r − w = tan( α/
2) tan( θ / w . Similarly, 1 r − w = tan( α/
2) tan( θ / w . But tan( θ / w = tan( θ / w by Theorem 4.2 (which will be proved in the next section). Thus,1 r − w = 1 r − w and the result follows. (cid:3) Theorem 3.4.
Chords AB and CD of a circle meet at E , with ∠ AEC = α . Let W ( w ) be the incircle of skewed sector BED and let W ( w ) be the incircle ofskewed sector AEC . Let O ( r ) be the incircle of (cid:52) BED and let O ( r ) be theincircle of (cid:52) AEC . See Figure 9. Then r r = w w cos α . Proof.
Let m ( (cid:95) DB ) = θ and m ( (cid:95) CA ) = θ . Applying Theorem 2.2 to skewedsector BED gives w r = cos( θ / α/
2) cos( θ / . Applying Theorem 2.2 to skewed sector
AEC gives w r = cos( θ / α/
2) cos( θ / . Multiplying these two equations gives w w r r = 1cos ( α/ tanley Rabinowitz and the result follows by cross-multiplying. (cid:3) Theorem 3.5.
Cevians AD and CF in (cid:52) ABC meet at P and ∠ BF C = ∠ BDA .The cevians are extended to meet the circumcircle of (cid:52)
ABC at points D (cid:48) and F (cid:48) ,respectively, as shown in Figure 10. Let W ( w ) be the incircle of skewed sector BDD (cid:48) and let W ( w ) be the incircle of skewed sector BF F (cid:48) . Let O ( r ) be theincircle of (cid:52) BDD (cid:48) and let O ( r ) be the incircle of (cid:52) BF F (cid:48) . Then w r = w r . Figure 10. w /r = w /r Proof.
Let m ( (cid:95) BD (cid:48) ) = θ , m ( (cid:95) F (cid:48) B ) = θ , and m ( (cid:95) CA ) = φ . Applying Theorem 2.2to skewed sector BDD (cid:48) using α = ∠ BDD (cid:48) gives w r = cos( φ/ α /
2) cos( θ / . Applying Theorem 2.2 to skewed sector
BF F (cid:48) using α = ∠ BF F (cid:48) gives w r = cos( φ/ α /
2) cos( θ / . But α = α because they are supplementary to the two given angles. Chords AD (cid:48) and BC intercept arcs of measures θ and φ , so α = ( θ + φ ) /
2. Similarly, α = ( θ + φ ) /
2. Thus, θ = θ because α = α . Therefore, w /r = w /r . (cid:3) Theorem 3.6.
Let H be the orthocenter of acute triangle ABC . The altitudes AD and CF are extended to meet the circumcircle of (cid:52) ABC at points D (cid:48) and F (cid:48) ,respectively. Let W ( w ) be the incircle of skewed sector BDD (cid:48) and let W ( w ) bethe incircle of skewed sector BF F (cid:48) . Let O ( r ) be the incircle of (cid:52) BDH and let O ( r ) be the incircle of (cid:52) BF H . See Figure 11. Then w r = w r . Relationships Between Circles Inscribed in Triangles and Curvilinear Triangles
Figure 11. w /r = w /r Proof.
Let r (cid:48) be the inradius of (cid:52) BDD (cid:48) and let r (cid:48) be the inradius of (cid:52) BF F (cid:48) .Since ∠ BF C = ∠ BDA , by Theorem 3.5, we have w /r (cid:48) = w /r (cid:48) . Now ∠ CBD (cid:48) = ∠ CAD (cid:48) since both angles subtend the same arc. But ∠ CAD (cid:48) = ∠ CBE (cid:48) since bothangles are complementary to ∠ ACB . Thus, ∠ DBD (cid:48) = ∠ DBH . Right triangles
BDD (cid:48) and
BDH share a common side. Thus (cid:52)
BDD (cid:48) ∼ = (cid:52) BDH . Hence r = r (cid:48) .Similarly, r = r (cid:48) . Therefore, w /r = w /r . (cid:3) Lemma 3.7.
Let H be the orthocenter of acute (cid:52) ABC , and let the altitudes be AD , BE , and CF as shown in Figure 12. Circles O ( r ) , O ( r ) , O ( r ) , and O ( r ) are inscribed in triangles BHD , BHF , CAF , and
ACD , respectively.Then r /r = r /r . Figure 12. r /r = r /r Proof.
Note that (cid:52)
BHF ∼ (cid:52)
CAF . Therefore the figure consisting of (cid:52)
BHF and its incircle is similar to the figure consisting of (cid:52)
CAF and its incircle. Cor-responding parts of similar figures are in proportion, so r /r = BH/CA . In thesame manner, (cid:52)
BHD ∼ (cid:52)
ACD which implies that r /r = BH/AC . Therefore, r /r = r /r or r /r = r /r . (cid:3) Corollary 3.8.
Let H be the orthocenter of acute (cid:52) ABC . The altitudes AD and CF are extended to meet the circumcircle of (cid:52) ABC at points D (cid:48) and F (cid:48) , tanley Rabinowitz respectively. Let W ( w ) be the incircle of skewed sector BDD (cid:48) and let W ( w ) bethe incircle of skewed sector BF F (cid:48) . Let O ( r ) be the incircle of (cid:52) AF C and let O ( r ) be the incircle of (cid:52) ADC . See Figure 13. Then w /w = r /r . Figure 13. w /w = r /r Proof.
By Theorem 3.6, w /w = r /r . By Lemma 3.7, r /r = r /r . Therefore, w /w = r /r . (cid:3) Relationships Between the Incircles of Two Skewed Sectors
Theorem 4.1.
Chords AB and CD of a circle meet at E . Let W ( w ) be the circleinscribed in skewed sector DEB and let W ( w ) be circle inscribed in skewed sector AEC , as shown in Figure 14. Then w w = tan( θ / θ / where ∠ BCD = θ and ∠ ADC = θ . Figure 14. w /w = tan( θ / / tan( θ / Proof.
See [4, pp. 96–97] or [11, p. 26–27]. (cid:3) Relationships Between Circles Inscribed in Triangles and Curvilinear Triangles
Since the measure of an angle inscribed in a circle is half the circular measure ofthe intercepted arc, we have the following result.
Theorem 4.2.
Chords AB and CD of a circle meet at E . Let W ( w ) be thecircle inscribed in skewed sector DEB and let W ( w ) be the circle inscribed inskewed sector AEC , as shown in Figure 15. Then w w = tan( θ / θ / where m ( (cid:95) CA ) = θ and m ( (cid:95) DB ) = θ . Figure 15. w /w = tan( θ / / tan( θ / θ /
4. If M is the midpoint of arc (cid:95) DB , then ∠ BDM = (cid:95) M B and (cid:95)
M B = (cid:95) DB = θ , so ∠ BDM = θ / Theorem 4.3.
Chords B B and C C of a circle meet at A . Let r and r bethe inradii of (cid:52) B AC and (cid:52) B AC , respectively, as shown in Figure 16. Let B C = a and let B C = a . Then = r /r = a /a . Figure 16. r /r = a /a Proof.
This follows from the fact that (cid:52) B AC ∼ (cid:52) B AC . (cid:3) The following theorem comes from [11, Problem 21] and is related to Ajima’sTheorem. tanley Rabinowitz Theorem 4.4.
Chords B B and C C of a circle meet at A . Let W ( w ) be thecircle inscribed in skewed sector B AC and let W ( w ) be the circle inscribed inskewed sector B AC . Let v and v be the heights of the segments formed bychords B C and B C as shown in Figure 17. Then w w = v a v a where a = B C and a = B C . Figure 17. w /w = v a /v a The following result is due to Pohoatza and Ehrmann, [6].
Theorem 4.5.
Let D be the point on side BC of (cid:52) ABC such that AB + BD = AC + CD . A circle is circumscribed about (cid:52) ABC . Let W ( w ) be the circleinscribed in skewed sector ADB and let W ( w ) be the circle inscribed in skewedsector ADC (Figure 18). Then w = w . Figure 18. w = w Proof.
Extend AD to meet the circumcircle of (cid:52) ABC at D (cid:48) . Let O ( r ) bethe circle inscribed in (cid:52) BDD (cid:48) and let O ( r ) be the circle inscribed in (cid:52) CDD (cid:48) (Figure 19). Then 1 /r + 1 /w = 1 /r + 1 /w by Theorem 3.2 (with points A and D (cid:48) interchanged). But r = r by Theorem 3.4 of [8]. Therefore, w = w . (cid:3) See [1] for another proof. Relationships Between Circles Inscribed in Triangles and Curvilinear Triangles
Figure 19. r = r and w = w Relationships Between the Incircles of Six Skewed Sectors
Theorem 5.1.
Let H be the orthocenter of acute (cid:52) ABC . The altitudes through H extended to meet the circumcircle of (cid:52) ABC divide the interior of that circumcircleinto six skewed sectors, each with vertex at H , as shown in Figure 20. Let W i ( w i ) be the circle tangent to two altitudes and internally tangent to the circumcircle asshown. Then w w w = w w w . Figure 20. w w w = w w w Proof.
Let θ i be the arc angle of the skewed sector containing circle W i . By The-orem 4.2, w /w = tan( θ / / tan( θ / w /w = tan( θ / / tan( θ / w /w = tan( θ / / tan( θ / w w w w w w = w w · w w · w w = tan( θ / θ / · tan( θ / θ / · tan( θ / θ / . Note that ∠ BAH = ∠ BCH since both are complementary to ∠ ABC . Therefore, θ = θ . Similarly, θ = θ and θ = θ . Hence w w w w w w = tan( θ / θ / · tan( θ / θ / · tan( θ / θ /
4) = 1 , so w w w = w w w . (cid:3) tanley Rabinowitz Theorem 5.2.
Let I be the incenter of (cid:52) ABC . The cevians through I extendedto meet the circumcircle of (cid:52) ABC divide the interior of that circumcircle into sixskewed sectors, each with vertex at I , as shown in Figure 21. Let W i ( w i ) be thecircle tangent to two cevians and internally tangent to the circumcircle as shown.Then w w w = w w w . Figure 21. w w w = w w w Proof.
Let θ i be the arc angle of the skewed sector containing circle W i ( w i ). ByTheorem 4.2, w /w = tan( θ / / tan( θ / w /w = tan( θ / / tan( θ / w /w = tan( θ / / tan( θ / w w w w w w = w w · w w · w w = tan( θ / θ / · tan( θ / θ / · tan( θ / θ / . Note that ∠ BAI = ∠ CAI since AI is an angle bisector. Therefore, θ = θ .Similarly, θ = θ and θ = θ . Hence w w w w w w = tan( θ / θ / · tan( θ / θ / · tan( θ / θ /
4) = 1 , so w w w = w w w . (cid:3) Theorem 5.3.
Let H be the orthocenter of acute (cid:52) ABC . The altitudes through H extended to meet the circumcircle of (cid:52) ABC divide the segments of the circumcirclebounded by the sides of the triangle into two skewed sectors each as shown inFigure 22. Let W i ( w i ) be the incircles of the six skewed sectors formed, situatedas shown in Figure 22. Then w w w = w w w .Proof. The altitudes of (cid:52)
ABC divide it into six triangles named T through T as shown in Figure 22. Let r i be the inradius of triangle T i . By Theorem 3.6, w /r = w /r . Similarly, w /r = w /r and w /r = w /r . Therefore, w w w r r r = w w w r r r . But r r r = r r r by Theorem 3.1 of [7]. Therefore, w w w = w w w . (cid:3) Relationships Between Circles Inscribed in Triangles and Curvilinear Triangles
Figure 22. w w w = w w w Theorem 5.4.
Let M be the centroid of (cid:52) ABC . The medians through M ex-tended to meet the circumcircle of (cid:52) ABC divide the segments of the circumcirclebounded by the sides of the triangle into two skewed sectors each as shown in Fig-ure 23. Let W i ( w i ) be the incircles of the six skewed sectors formed, situated asshown in Figure 23. Then w + 1 w + 1 w = 1 w + 1 w + 1 w . Figure 23. /w + 1 /w + 1 /w = 1 /w + 1 /w + 1 /w Proof.
A cevian through a point P inside a triangle ABC divides (cid:52)
ABC intotwo triangles, known as side triangles . There are six such side triangles, named S through S as shown in Figure 24. Figure 24. naming of side triangles tanley Rabinowitz Let r i be the radius of the circle inscribed in triangle S i . When P is the centroidof (cid:52) ABC , Theorem 2.2 from [8] states that1 r + 1 r + 1 r = 1 r + 1 r + 1 r . By Theorem 3.2, (cid:18) w − w (cid:19) + (cid:18) w − w (cid:19) + (cid:18) w − w (cid:19) = (cid:18) r − r (cid:19) + (cid:18) r − r (cid:19) + (cid:18) r − r (cid:19) = 0 , so 1 /w + 1 /w + 1 /w = 1 /w + 1 /w + 1 /w . (cid:3) Theorem 5.5.
Let H be the orthocenter of acute (cid:52) ABC . The altitudes through H divide the triangle into six side triangles, S through S as shown in Figure 24.Let W i ( w i ) be the incircle of the skewed sector associated with S i . Two of thesecircles are shown in Figure 25. Then w w w = w w w . Figure 25. w w w = w w w Proof.
This follows from Theorem 5.3 by applying Corollary 3.8. (cid:3)
Theorem 5.6.
Let O be the circumcenter of (cid:52) ABC . The cevians through O extended to meet the circumcircle of (cid:52) ABC divide the interior of that circumcircleinto six skewed sectors, each having vertex at O , as shown in Figure 26. Let W i ( w i ) be the circle tangent to two cevians and internally tangent to the circumcircle asshown. Then w = w , w = w , w = w . Proof.
It suffices to show that w = w . Note that ∠ BOD (cid:48) = ∠ AOE (cid:48) becausethey are vertical angles. Also, OB = OD (cid:48) = OE (cid:48) = OA because they are all radiiof circle O . Therefore, skewed sectors BOD (cid:48) and
AOE (cid:48) are congruent and thustheir incircles are also congruent. (cid:3)
Theorem 5.7.
Let O be the circumcenter of (cid:52) ABC . The cevians through O areextended to meet the circumcircle of (cid:52) ABC at points D (cid:48) , E (cid:48) , and F (cid:48) as shownin Figure 27. The cevians divide (cid:52) ABC into six side triangles named S through S as shown in Figure 24. Six circles, W i ( w i ) , are inscribed in the skewed sectors Relationships Between Circles Inscribed in Triangles and Curvilinear Triangles
Figure 26. w = w , w = w , w = w associated with these side triangles. Two of these circles are shown in Figure 27.Then w = w , w = w , w = w . Figure 27. w = w Proof.
It suffices to show that w = w . Note that OA = OB because they areboth radii of circle O . Thus, ∠ OAB = ∠ OBA because they are the base angles ofan isosceles triangle. Also, AD (cid:48) = BE (cid:48) because they are both diameters of circle O . The skewed sectors BAD (cid:48) and
ABE (cid:48) have side AB in common. Therefore,skewed sectors BAD (cid:48) and
ABE (cid:48) are congruent and hence their incircles are alsocongruent. (cid:3)
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