INAMO Shortlist 2015 G4 wrote:
Given an isosceles triangle $ABC$ with $AB = AC$, suppose $D$ is the midpoint of the $AC$. The circumcircle of the $DBC$ triangle intersects the altitude from $A$ at point $E$ inside the triangle $ABC$, and the circumcircle of the triangle $AEB$ cuts the side $BD$ at point $F$. If $CF$ cuts $AE$ at point $G$, prove that $AE = EG$.
Solution:-
Claim :-$\triangle AED\cong\triangle FED$.
For this note that $\angle EBF=\angle ECA=\angle EBA\implies EA=EF$. Also $\angle EFD=\angle BAE=\angle DAE$ and $\angle EDB=\angle ECB=\angle EBC=\angle ADE$. So ,due to the congruence of the triangles $AED$ and $FED$ we get that $AD=DF=DC$. Also $EB=EC$ and $\angle BAC=\angle FDC$. So now due to the Similarity of the triangles $BEC$ and $FDC$ we get that $\angle DFC=\angle EBC=\angle ECB=\angle EDF\implies ED\|FC\implies E$ is the midpoint of $AG$, in other words. $AE=EG$. $\blacksquare$.