Iranian Mathematical Olympiad

Denote by $\Gamma$ the circumcircle of the triangle $ABC$. Letting $F$ and $G$ be the reflections of $H$ with respect to $AC$ and $AB$, respectively. It is well-known that $F$ and $G$ lie on $\Gamma$. Also $EF$ is parallel to $AC$ since $A$ is the midpoint of the segment $EH$. Therefore $\angle EFB=90^{\circ }$ and the pentagon $EFDBX$ is cyclic. Now notice that $$\angle XFB=\angle XDB=\angle ACB=\angle AFB,$$

Yielding that the line $FX$ passes through $A$. Then, since $AF=AH=AE$ we should have $AD=AX$. Similarly one can show that $G$, $A$, and $Y$ are collinear and $AD=AY$. So $AX=AY$ and $AF=AG$. This yields that $XY\parallel FG$ and $$\angle XAG=\angle AYX+\angle AXY=\angle AYX+\angle AFG.$$ Hence the result follows. ■