Let $\Gamma$ be the circumcircle of an acute triangle $ABC$. The perpendicular line to $AB$ passing through $C$ cuts $AB$ in $D$ and $\Gamma$ again in $E$. The bisector of the angle $C$ cuts $AB$ in $F$ and $\Gamma$ again in $G$. The line $GD$ meets $\Gamma$ again at $H$ and the line $HF$ meets $\Gamma$ again at $I$. Prove that $AI = EB$.
Problem
Source: 2018 Pan-African Shortlist - G6
Tags: geometry, angle bisector
Zhang2018
07.05.2019 16:49
DylanN, how did you put the source of the problem at the top of the thread? @below, thanks. I usually post in MSM, so I did not see the source tool. The source function is present in the "High School Olympiads" forum, but it does not seem to be present in MSM.
DylanN
07.05.2019 16:52
When you add a new post, there is a line for the source just below the line for the tags.
Vlados021
07.05.2019 17:07
Pascal to $CGGHIE$, and q.e.d.
zuss77
09.05.2019 16:00
No need in $CD \perp AB$. It's South Africa 2017 P5
EulersTurban
07.12.2020 02:16
Our end goal is to prove that $CI$ and $CE$ are isogonal. Thus this means $C,O,I$ are colinear points. Let $T$ be a point on the circumcircle such that $CE$ and $CT$ are isogonal. Throw the configuration onto the complex plane such that $b=\frac{1}{a}$. This gives us that $g=-1$. We easily have that $t=-c$. By simple calculation we have that: $$h=\frac{a+d}{1+\overline{d}}=\frac{2+a+\frac{1}{a}+c-\frac{1}{c}}{2+a+\frac{1}{a}+\frac{1}{c}-c}$$ By simple calculation we easily have that: $$f=\frac{a^2c+c+ac-a}{a(c+1)}$$ By calculating we have that: $$\frac{h-f}{\overline{h-f}}=\frac{h-t}{\overline{h-t}}$$ Thus this implies that $H,F,T$ are colinear this means that $T\equiv I$, thus we are done.