131. A stronger Form of the Steiner-Lehmus Theorem (own).

by Virgil Nicula, Sep 25, 2010, 2:40 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=156786&hilit=bisector

Let $ABC$ be a triangle and let $E\in (AC)$ , $F\in (AB)$ so that $\widehat{EBA}\equiv\widehat{EBC}$ , $\widehat{FCA}\equiv\widehat {ECB}$ . Prove that $BE=CF\ \iff\ AB=AC$ .

Proof 1 (metric). $BE=CF$ $\iff$ $\frac {2}{a+c}\cdot\sqrt {acs(s-b)}= \frac {2}{a+b}\cdot\sqrt {abs(s-c)}$ $\iff$ $b(s-c)(a+c)^2=c(s-b)(a+b)^2$ $\iff$

$s\left[b(a+c)^2-c(a+b)^2\right]=bc\left[(a+c)^2-(a+b)^2\right]$ $\iff$ $s\left[b(a^2+c^2)-c(a^2+b^2)\right]=bc(2s+a)(c-b)$ $\iff$

$s(bc-a^2)(c-b)=bc(2s+a)(c-b)$ $\iff$ $(b-c)\left[bc(2s+a)-s(bc-a^2)\right]=0$ $\iff$ $(b-c)(sbc+abc+sa^2)=0$ $\iff$ $b=c$ .

Proof 2 (metric and reductio ad absurdum). $B>C\ \iff\ \left\|\begin{array}{ccc} b>c & \iff & \frac {2ac}{a+c}<\frac {2ab}{a+b}\\\\ \frac B2>\frac C2 & \iff & \cos\frac B2<\cos\frac C2\end{array}\right\|$ $\implies$ $\frac {2ac}{a+c}\cdot\cos\frac B2<\frac {2ab}{a+b}\cdot\cos\frac C2$ $\iff$

$BE<CF$ , absurd. Prove analogously that $B<C\ \implies\ BE>CF$, absurd. Thus, $BE=CF\implies B=C$ . Prove easily $B=C\implies BE=CF$ .

Proof 3 (synthetical and reductio ad absurdum). Construct the parallelogram $BEGF$ ($EG\parallel BF$ and $FG\parallel BE$) . Since $FG=BE=CF$ obtain $FG=FC$ , i.e. the triangle $CFG$ is $F$-isosceles and $\widehat{FGC}\equiv\widehat{FCG}$ . Denote $x=m(\widehat {ABE})=m(\widehat{CBE})$ , $y=m(\widehat{BCF})=m(\widehat{ACF})$ and $x'=m(\widehat{EGC})$ , $y'=m(\widehat{ACG})$ . Since $\widehat{FGC}\equiv\widehat{FCG}$ obtain $x+x'=y+y'$ . Suppose contrary that $B>C$ , i.e. $2x>2y$ $\iff$ $x>y$ $\iff$ $x'<y'$ $\iff$ $EC<EG$ $\iff$ $\underline{EC<FB}$ . The triangles $BCF$ , $CBE$ have a common side $[BC]$ and $CF=BE$ . From the relation $m(\widehat{CBE})=x>y=m(\widehat{BCF})$ obtain $\underline{EC>FB}$ , what is absurd. Prove analogously $\underline{B<C\implies EC<FB}$ , what is absurd. In conclusion, $B=C$ , i.e. $AB=AC$ .

Proof 4 (synthetical). Let $H$ be the point so that $\triangle BAE\equiv\triangle FHC$ and $CF$ separates $B$ , $H$ . Denote $J\in (CF)$ for which $\widehat{JHC}\equiv\widehat{JHF}$ $\implies$ $\widehat{BAE}\equiv \widehat{FHC}$

i.e. $\widehat{FAC}\equiv\widehat{FHC}$ $\implies$ $AFCH$ is cyclically. Thus, $\widehat{ACF}\equiv\widehat{AHF}$ , $\widehat{BAI}\equiv\widehat{IAC}\equiv\widehat{FHJ}\equiv\widehat{CHJ}$ . Since $AIJH$ is cyclically and $AI=HJ$

obtain $IJ\parallel AH$ . In conclusion, $\widehat{ABI}\equiv\widehat{HFC}\equiv\widehat{FHA}\equiv\widehat{FCA}$ , i.e. $\widehat{IBC}\equiv\widehat{ICB}$ , what means $\triangle BIC$ is $I$-isosceles $\implies$ $\triangle ABC$ is $A$-isosceles.

Extension. Let $ABC$ be a triangle and let $D\in (BC)$ be the point for which $\widehat{DAB}\equiv\widehat{DAC}$ . For a point

$L\in (AD)$ denote $E\in BL\cap AC$ , $F\in CL\cap AB$ . Prove that $BE=CF\ \iff\ AB=AC$ .

Proof 1 (metric). Denote $\left\|\begin{array}{c} \frac {EA}{x}=\frac {EC}{1}=\frac {b}{x+1}\\\\ \frac {FA}{y}=\frac {FB}{1}=\frac {c}{y+1}\end{array}\right\|$ . Apply the Ceva's teorem to $L$ in $\triangle\ ABC\ :\ \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB}=1$ $\iff$ $\boxed{bx=cy}\ (1)$ .

Apply the Stewart's theorem for the cevians $BE$ , $CF\ :\ \left\|\begin{array}{c} (x+1)^2\cdot BE^2=(a^2x+c^2)(x+1)-b^2x=a^2x(x+1)+c^2(x+1)-bcy\\\\ (y+1)^2\cdot CF^2=(a^2y+b^2)(y+1)-c^2y=a^2y(y+1)+b^2(y+1)-bcx\end{array}\right\|$ .

Thus, $BE=CF\iff$ $(y+1)^2\left[a^2x(x+1)+c^2(x+1)-bcy\right]\stackrel{(1)}{=}(x+1)^2\left[a^2y(y+1)+b^2(y+1)-bcx\right]$ $\iff$

$a^2(x+1)(y+1)\left[x(y+1)-y(x+1)\right]+$ $bc\left[x(x+1)^2-y(y+1)^2\right]+(x+1)(y+1)\left[c^2(y+1)-b^2(x+1)\right]=0$ . Observe that

$\left\|\begin{array}{c} x(y+1)-y(x+1)=x-y\\\\ x(x+1)^2-y(x+1)^2=(x-y)\left(x^2+xy+y^2+2x+2y+1\right)\\\\ c^2(y+1)-b^2(x+1)\stackrel{(1)}{=}bc(x-y)\left(\frac {x+y}{xy}+1\right)\end{array}\right\|$ . Therefore, $BE=CF\iff$

$\boxed{(x-y)\left\{a^2(x+1)(y+1)+bc\left[\left(x^2+xy+y^2+2x+2y+1\right)+(x+1)(y+1)\left(\frac {x+y}{xy}+1\right)\right]\right\}=0}$ $\iff$

$x=y$ $\iff$ $\frac {EA}{EC}=\frac {FA}{FB}\iff$ $EF\parallel BC$ $\iff$ $AL$ is $A$-median ($AL$ is and $A$-bisector) in $\triangle\ ABC$ $\iff\triangle\ ABC$ is $A$-isoceles $\iff$ $AB=AC$ .

Proof 2 (metric). Denote $\left\|\begin{array}{c} \frac {EA}{x}=\frac {EC}{1}=\frac {b}{x+1}\\\\ \frac {FA}{y}=\frac {FB}{1}=\frac {c}{y+1}\end{array}\right\|$ . Apply the Ceva's theorem to $L$ in $\triangle\ ABC\ :\ \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB}=1$ $\iff$ $\boxed{\frac by=\frac cx=\lambda}\ (1)$ .

Pytagoras' theorem to $[BE]$ , $[CF]$ in $\triangle BCE$ , $\triangle BCF$ $\iff$ $\left\{\begin{array}{c} (x+1)^2\cdot BE^2=(x+1)^2a^2+b^2-(x+1)(a^2+b^2-c^2)\ .\\\\ (y+1)^2\cdot CF^2=(y+1)^2a^2+c^2-(y+1)(a^2+c^2-b^2)\ .\end{array}\right\|$ .

Thus $CE=BF\iff$ $(y+1)^2\left[(x+1)^2a^2+b^2-(x+1)\left(a^2+b^2-c^2\right)\right]=$ $(x+1)^2\left[(y+1)^2a^2+c^2-(y+1)\left(a^2+c^2-b^2\right)\right]$ $\iff$

$(x-y)(x+1)(y+1)a^2+$ $(y+1)\left[(y+1)-(x+1)(y+1)-(x+1)^2\right]b^2+$ $(x+1)\left[(y+1)^2+(x+1)(y+1)-(x+1)\right]c^2=0$ $\stackrel{(1)}{\iff}$

$(x-y)(x+1)(y+1)a^2+$ $\lambda^2\left[x^2(x+1)(y^2+xy+3y+1)-y^2(y+1)(x^2+xy+3x+1)\right]=0$ $\iff$

$\boxed{(x-y)\left\{(x+1)(y+1)a^2+\lambda^2\left[2x^2y^2+xy(x^2+y^2)+4xy(x+y)+4xy+(x^2+y^2)+(x+y)\right]\right\}=0}$ $\iff$ $x=y$ a.s.o.

Proof 3 (synthetical). Let $H$ be the point so that $\triangle FHC\equiv\triangle BAE$ and the line $CF$ separates $B$ , $H$ . Denote the point $J\in (CF)$ for which $\widehat{JHC}\equiv\widehat{JHF}$ .

Therefore, $\widehat{FAC}\equiv \widehat{FHC}$ $\implies$ $AFCH$ is cyclically. Thus, $\widehat{CAH}\equiv\widehat{CFH}\equiv\widehat {EBA}$ . Observe that $m(\widehat {CJH})=$ $m(\widehat {ELA})=$

$m(\widehat{BAL})+m(\widehat{EBA})=$ $m(\widehat{LAC})+m(\widehat{CAH})=$ $m(\widehat{LAH})$ . Therefore $\widehat {CJH}\equiv\widehat{LAH}$ , i.e. $ALJH$ is cyclically. From $AL=HJ$ obtain that $ALJH$

is isosceles trapezoid and $IJ\parallel AH$ , i.e. $CF\parallel AH$ . Thus, $\widehat{CAH}\equiv\widehat{ACF}$ $\implies$ $\widehat{EBA}\equiv\widehat{ACF}$ . In conclusion, $\triangle ABE\equiv\triangle ACF$ , i.e. $AB=AC$ .


Generalization. Let $ABC$ be a triangle and let $D\in (BC)$ be the intersection of $A$-symmedian with the sideline $BC$ .

For a point $L\in (AD)$ denote $E\in BL\cap AC$ , $F\in CL\cap AB$ . Prove that $BE=CF\ \iff\ AB=AC$ .

Proof (metric). Denote $\left\|\begin{array}{c} \frac {EA}{x}=\frac {EC}{1}=\frac {b}{x+1}\\\\ \frac {FA}{y}=\frac {FB}{1}=\frac {c}{y+1}\end{array}\right\|$ . Apply the Ceva's theorem to $L$ in $\triangle\ ABC\ :\ \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB}=1$ $\iff$ $\boxed{\frac {b^2}{y}=\frac {c^2}{x}=\lambda}\ (1)$ .

Pytagoras' theorem to $[BE]$ , $[CF]$ in $\triangle BCE$ , $\triangle BCF$ $\iff$ $\left\{\begin{array}{c} (x+1)^2\cdot BE^2=(x+1)^2a^2+b^2-(x+1)(a^2+b^2-c^2)\ .\\\\ (y+1)^2\cdot CF^2=(y+1)^2a^2+c^2-(y+1)(a^2+c^2-b^2)\ .\end{array}\right\|$ .

Thus $CE=BF\iff$ $(y+1)^2\left[(x+1)^2a^2+b^2-(x+1)\left(a^2+b^2-c^2\right)\right]=$ $(x+1)^2\left[(y+1)^2a^2+c^2-(y+1)\left(a^2+c^2-b^2\right)\right]$ $\iff$

$(x-y)(x+1)(y+1)a^2+$ $(y+1)\left[(y+1)-(x+1)(y+1)-(x+1)^2\right]b^2+$ $(x+1)\left[(y+1)^2+(x+1)(y+1)-(x+1)\right]c^2=0$ $\stackrel{(1)}{\iff}$

$(x-y)(x+1)(y+1)a^2+$ $\lambda\left[x(x+1)(y^2+xy+3y+1)-y(y+1)(x^2+xy+3x+1)\right]=0$ $\iff$

$(x-y)\left\{(x+1)(y+1)a^2+\lambda\left[2xy+(xy+1)(x+y+1)\right]\right\}=0$ $\iff$ $x=y$ a.s.o.
This post has been edited 83 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:26 AM

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143. Assess of a ratio (nice and difficult problem).
142. Inscribed/circumscribed quadrilateral w.r.t. ABC.
141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).
140. A synthetical property <==> linear/angled relation.
139. A caracterization of a right angle in a triangle.
138. Generalized "slicing" problems (own).
137. Stewart's relation.
136. A geometrical interpretation of a system (own).
135. Three strong geometrical inequalities.
+ September 2010
134. Some problems with projections/perpendiculars (2).
133. Some problems with projections/perpendiculars (1).
132. Asupra unei inegalitati intr-un triunghi.
131. A stronger Form of the Steiner-Lehmus Theorem (own).
130. A triangle and a fixed point.
129. BX=CY and three properties.
128. Value of IG and two inequalities in neighbourhood of 0.
127. Pagina educativa.
126. Routh's theorem.
125. I like very much this Darij Grinberg's message.
124. A nice and easy problem with a right-angled triangle.
123. My extension and its proof.
122. An identity for an equilateral triangle. Extension.
121. Viete's relations with cos A , cos B , cos C.
120. Two equal neighbouring angles/sides in a quadrilateral.
119. Inspired by a Miculita's problem.
118. An usual remarkable geometrical locus.
117. Some applications of harmonic division/quadrilateral.
116. The length of AE (nine methods !).
115. An difficult equation with one radical.
114. A property of tangential triangle for ABC.
113. Range for a function.
112. The equivalence relation ".s.s."
111. Simple, but interesting ...
110. German TST 2004 - Exam V , Problem 1
109. Two equal areas ==> find A.
108. Old, short and nice proof of Euler's relation.
107. XY parallel BC and X,Y are isogonal/izotomic points.
106. OLM - Bucuresti (geometrie).
105. A problem with three circles.
104. Concursul Revistei de Matematica din Galati (RMG).
103. Metrical usual relations in triangle. Applications.
102. Characterization of a metrical relation in a triangle.
101. O problema cu numere complexe.
100. Bobiller's theorem.
99. Some simple and nice geometry problems.
98. Symmedian & median (metrical relations).
97. Problem of butterfly.
+ August 2010
96. Integrals.
95. Three circles in an angle.
94. Equations of angle bisectors (interior and exterior).
93. The Appolonius' construct problems.
92. CA , CB tangent to parabola - OIMU 2008 Problem 4.
91. JBMO 2010, Problem 3.
90. Nice and difficult inequality in ABC (own).
89. Similarity (parallel/antiparallel).
88. A nice geometric locus conditioned metrically.
87. An interesting vectorial identity.
86. A very nice maxmin.
85. Problem "slicing" : 60-24-12.
84. IRAN national math olympiad - 2010
83. A metrical characterization of concurrence.
82. AA' , BN , CM concur in Gergonne point.
81. Sawayama and Thebault’s theorem.
80. Three concurrent perpendiculars (the Carnot's lemma).
79. A quadrilateral with many perpendicularities.
78. Jakobi's theorem.
77. Two equivalent perpendicularities.
76. A parallelogram and a circle (nice !).
75. Extended Steinbart Theorem.
74. Inegalitatea Erdos-Mordell.
73. A nice problems with a square and a perpendicularity.
+ July 2010
72. Colinearity in a triangle - H , P , M .
70. Colinearity with pole/polar - H\in PQ .
71. An angle questions (synthetic/trigonometric proofs).
69*1. Slicing problem 3.
69. Position of x1, x2 w.r.t. a given real number k.
68. Some nice applications of "pole/polar".
67. Remarkable algebraic/geometrical inequalities (1).
66. Remarkable perpendicularities in a triangle.
65. Parallel to BC through I .
64. Problems of the type "instant" (at most one minute).
63. A nice problem with radical center (livetolove212).
62. Nice and easy problem with a right triangle.
61. IMO Shortlist 2009 (very nice problem and its proof).
60. Mediteranean M.C. 2010 Problem 3.
59. IMO 2010, problem 4.
58. A cevian and two incircles.
57. Opinii si comentarii relativ la inv. rom. II
56. A extremum problem with areas.
55. Two important circles - mixtilinear incircle/excircle.
54. Cinci grade de libertate.
53. An remarkable concurrency.
52. Harmonical division.
51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.
51. A nice problem with perpendicularity from Jaime.
50. IMAC 2010 seniors, the first day.
49. The midpoint of a cevian in an acute triangle.
+ June 2010
48. An inequality in a triangle : h_a+max{b,c} ...
47. Interesting and difficult identities in a triangle.
46. Outside of a quadrilateral.
45. \cos B+\cos C=1 <==> nice property.
44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.
43. A "slicing" problem with a parameter.
42. IMAC 2010 Problema 2.
41. ONM 2010 Calarasi Problema 3.
+ May 2010
40. Lema lui Haruki.
39. A nice and remarkable problem with Brocard's point.
38. Minimum area of triangle touching with semicircle.
37. Own extension of the Steiner-Lehmus' problem.
36. ONM (juniori) - 2010 Iasi, problema 4.
35. Nice problem from Arqady (Eduard_Tur).
34. IMO Shortlist 2002 geometry problem 7.
+ April 2010
32. Linkuri diverse.
31.Some nice metrical problems.
29. Inegalitate integrala de la Kunny.
28. A fost odată ...
27.Problem of Darij Grinberg (I - centroid of triangle DEF).
26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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