Singapore Mathematical Olympiad (SMO)

Note that $A$, $F$, $G$, $B$ are concyclic, since they lie on the circle $\Gamma$ with diameter $AB$. Let $DF$ intersect $\Gamma$ again at $H^{\prime }$; it suffices to show that $H=H^{\prime }$.

Solution 1: Let the diagonals $AC$ and $BD$ intersect at $O$; note that $O$ also lies on $\Gamma$. Let $H^{\prime }G\cap AO=C^{\prime }$. Now Pascal's theorem on $FAOBG{H}^{\prime }$ shows that $C^{\prime }$, $D$, $E$ are collinear, so $C=C^{\prime }$ and $H=H^{\prime }$.

Solution 2: By power of point, $DF\times D{H}^{\prime }=D{A}^2=D{C}^2$. Therefore $△D{H}^{\prime }C\sim △DCF$. Therefore, since $\angle ECB=\angle EFB=90^{\circ }$ implies that $ECBF$ is cyclic, $$\angle D{H}^{\prime }C=\angle DCF=\angle ECF=\angle EBF=\angle EBF$$ Hence if $C{H}^{\prime }$, $EB$ intersect at $G^{\prime }$ then $F{G}^{\prime }B{H}^{\prime }$ is cyclic, which implies $G^{\prime }$ lies on $\Gamma$, so $G^{\prime }=G$ and thus $H^{\prime }=H$.