Iranian Mathematical Olympiad

Let $\Omega$ be the circumcircle of $DXF$ and $l$ be the perpendicular bisector of $FX$. Note that lines $AB$, $CD$ are external common tangents to $\Omega$, $w$, therefore they are reflection of each other with respect to $l$. Suppose that $AB$ is tangent to $\Omega$ at $Y$, and $Z$ is the reflection of $A$ with respect to $l$. The line $XY$ is the reflection of $FD$ with respect to $l$ hence $FD$ passes through $A$, and $XY$ passes through $Z$.

It is clear that $AFXZ$ is an isosceles trapezoid therefore $AFXZ$ is cyclic and it is suffices to prove that the circumcircle of $FXZ$ passes through $C$. Note that $\angle BEF=\angle FXE=\frac{\widehat{FXD}}{2}=\angle FDC$ therefore $FECD$ is cyclic and we have $$\angle FCZ=\angle FCD=\angle FED=\angle FEX=\angle FXY=180^{\circ }-\angle FXZ.$$ Which implies that $FXZC$ is cyclic.