Let $\Gamma$ be the circumcircle of an acute triangle $ABC.$ The perpendicular to $AB$ from $C$ meets $AB$ at $D$ and $\Gamma$ again at $E.$ The bisector of angle $C$ meets $AB$ at $F$ and $\Gamma$ again at $G.$ The line $GD$ meets $\Gamma$ again at $H$ and the line $HF$ meets $\Gamma$ again at $I.$ Prove that $AI = EB.$
Problem
Source: Baltic Way 2014, Problem 11
Tags: geometry, circumcircle, geometric transformation, reflection, geometry proposed
TelvCohl
11.11.2014 17:23
My solution: Redefine $ I $ as a point on $ \Gamma $ satisfy $ AI=BE $ and $ H=DG \cap FI $ . (ie. $ I $ is the reflection of $ C $ in the center of $ \Gamma $ ) We'll prove $ H \in \Gamma $ . Since $ \angle DCF=\angle GCI, \angle FDC=\angle IGC=90 $ , so $ \triangle CFD \sim \triangle CIG $ . ie. $ \triangle CDG \sim \triangle CFI $ hence we get $ \angle GHI=\angle GCI $ . ie. $ G, H, I, C $ are concyclic or $ H \in \Gamma $ Q.E.D
Jul
11.11.2014 17:29
Easily, we have : \[\Delta DAG\sim \Delta AHG\rightarrow GA^2=GD.GH\] \[\Delta FAG\sim \Delta ACG\rightarrow GA^2=GF.GC\] Implies $GD.GH=GF.GC$. So $HCFD$ is cyclic. Thus $\angle CHF=\angle CDF=90^0$. We can continue to solve this problem easily.
anantmudgal09
13.09.2015 02:37
$D$ can be any point on $BC$. By power of a point we get $H,F,D,C$ concyclic. Now angle chase.
rafaello
13.08.2021 23:57
Claim. $DFCH$ is cyclic. Proof. By Reim's theorem, the claim follows. Other way is to angle chase, $$\measuredangle DHC=\measuredangle (\ell,GC)=\measuredangle DFG=\measuredangle DFC,$$where $\ell$ is the tangent to $(ABC)$ from $G$. By the claim, $\measuredangle GHI=\measuredangle DHF=\measuredangle DCF=\measuredangle ECG$, thus $G$ is the midpoint of arc $EI$. As $G$ is the midpoint of arc $AB$ also, we get that $ABIE$ is isosceles trapezoid. [asy][asy] size(7cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,C,D,E,F,G,H,I,K; O=(0,0);A=dir(95);B=dir(200);C=dir(340);D=foot(C,A,B);path w=circumcircle(A,B,C); E=intersectionpoints(w,C--100D-99C)[0];K=incenter(A,B,C);F=extension(C,K,A,B);G=intersectionpoints(w,C--100F-99C)[0];H=intersectionpoints(w,G--100D-99G)[0];I=intersectionpoints(w,H--100F-99H)[1]; draw(A--B--C--cycle,deep);draw(w,deep);draw(C--E,light);draw(C--E,light);draw(C--G,light);draw(G--H,light);draw(H--I,light); draw(circumcircle(D,F,C),med+dashed); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$G$",G,dir(G)); dot("$H$",H,dir(H)); dot("$I$",I,dir(I)); [/asy][/asy]