Given a triangle $ABC$,with $I$ as its incenter and $\Gamma$ as its circumcircle,$AI$ intersects $\Gamma$ again
at $D$.Let $E$ be a point on the arc $BDC$, and $F$ a point on the segment $BC$,such that
$\angle BAE=\angle CAF<\frac {1}{2} \angle BAC$.If $G$ is the midpoint of $IF$,prove that the meeting point of
the lines $EI$ and $DG$ lies on $\Gamma$.
Remarks:
I was in waters to solve this problem,I had no idea what to do with $G$.
Then guess who saved me? Zhao!!(যার কেউ নাই তার ঝাও আছে ).
A nice homothety took $G$ to $F$ and $D$ to the $A$-excenter.Then some angle chasing,similar triangles and working with ratios solved the problem
আমি এটার যে Solution জানি সেটার আউটলাইন দিলাম। মুন ভাইয়ার করা।আমি কাছাকাছি গিয়েছিলাম। কিন্তু শেষ কোণটুকু বের করতে পারি নাই।
Let $DG\cap AF=X,DG\cap \tau =I$.
Now using Menelaus and sine rule, we can prove that $IX\parallel BC$. Now use the fact that $ATDC$ is concyclic,we can show that $ATXI$ is concyclic. Then we only need to show that $\angle BAF=\angle CAE'$. This is left for you!
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
).
