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Radical Axis and Circumcenter
azagorod101   5
N Jul 2, 2016 by Reynan
Source: BMO 2012/6
Let $ABC$ be a triangle. Let $S$ be the circle through B tangent to $CA$ at $A$ and let $T$ be the circle through C tangent to $AB$ at $A$. Circles $S$ and $T$ intersect at $A$ and $D$. Let $E$ be where $AD$ meets circle $ABC$. Prove that $D$ is the midpoint of $AE$.
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azagorod101
May 17, 2016
Reynan
Jul 2, 2016
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Radical Axis and Circumcenter
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geometry
power of a point
radical axis
circumcircle
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Source: BMO 2012/6
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azagorod101
194 posts
azagorod101
#1 h  May 17, 2016, 6:39 PM • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be a triangle. Let $S$ be the circle through B tangent to $CA$ at $A$ and let $T$ be the circle through C tangent to $AB$ at $A$. Circles $S$ and $T$ intersect at $A$ and $D$. Let $E$ be where $AD$ meets circle $ABC$. Prove that $D$ is the midpoint of $AE$.
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ThE-dArK-lOrD
4071 posts
ThE-dArK-lOrD
#2 h  May 17, 2016, 7:06 PM • 2 Y
Y by Adventure10, Mango247
Let $\angle{BAD}=x=\angle{ACD}=\angle{BCE},\angle{DAC}=y=\angle{ABD}=\angle{EBC}$
We need to prove $\frac{DE}{BD} \cdot \frac{BD}{AD} =1$ $\Leftrightarrow$ $\frac{sin(\angle{EBD})}{sin(\angle{BEA})} \cdot \frac{sin(x)}{sin(y)}=1$
Note that $\angle{EBD} =\angle{ABC}$ and $\angle{BEA} =\angle{ACB}$
So we need $\frac{AC}{AB} \cdot \frac{BE}{CE} =1$
Note that $\triangle{BEC} \sim \triangle{BDA}$ give $\frac{BE}{CE} =\frac{BD}{AD}$ and $\frac{BD}{AD} =\frac{AB}{AC}$ since $\triangle{BDA} \sim \triangle{ADC}$
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nikolapavlovic
1246 posts
nikolapavlovic
#3 h  May 17, 2016, 7:11 PM • 2 Y
Y by Adventure10, Mango247
$D$ is the center of spiral similarity sending $AC$ to $BA$ so $AD$ is the symmedian.Hence $BC$ is the symmedian in $\triangle EBA$ so we need to prove
$\angle EBC=\angle ACD$ but $\angle EBC=\angle CAE=\angle ACD$ by the above similarity
This post has been edited 2 times. Last edited by nikolapavlovic, May 17, 2016, 7:14 PM
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Ghost_rider
35 posts
Ghost_rider
#4 h  Jul 1, 2016, 9:55 PM • 2 Y
Y by Adventure10, Mango247
Let $AD$ $\cap$ $BC$ = $M$. Since $\bigtriangleup$ $BAD$ $\sim$ $\bigtriangleup$ $ADC$ ( $\frac{AB}{AD}$=$\frac{AC}{DC}$ , $\frac{AB}{BD}$=$\frac{AC}{AD}$) and $\angle$ $BDE$ = $\angle$ $EDC$ so $\frac{AB^2}{AC^2}$=$\frac{BM}{MC}$ $\longrightarrow$ $AM$ is symmedian. Let the tangent to $B$ and $C$ intersect in $N$ so $A$,$D$,$E$ and $N$ are collinear. Is well-know that $AB.EC$ = $BE.AC$ and $\frac{AM}{ME}$=$\frac{AB.AC}{BE.EC}$ so $\frac{AC^2}{CE^2}$=$\frac{AM}{ME}$ $\longrightarrow$ $MC$ is symmedian. Then $AD$ = $DE$.
This post has been edited 1 time. Last edited by Ghost_rider, Jul 1, 2016, 9:55 PM
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jayme
9774 posts
jayme
#5 h  Jul 2, 2016, 6:53 AM • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,

see

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=597823

Sincerely
Jean-Louis
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Reynan
632 posts
Reynan
#6 h  Jul 2, 2016, 7:41 AM • 2 Y
Y by Adventure10, Mango247
let $EB$ intersect $S$ again at $F$ and $EC$ intersect $T$ again at $G$. $EB\cdot EF= ED\cdot EA= EC\cdot EG$ makes $FBCG$ cyclic. $\angle ADF=\angle ABF=\angle ACE=180-\angle ACG=180 - \angle ADG$ makes $FDG$ collinear. $\angle AFD=\angle CAD = \angle CGD = \angle EGD$ makes $AF\parallel EG$. Analog $AG\parallel EF$. These make $AEFG$ parallelogram with centre $D$. DONE:)
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