Selection tests for the Gulf Mathematical Olympiad 2013

We have $\angle AFG=\angle ADG$ since $AGDF$ is cyclic. On the other hand $\angle DGA=\angle AFD=90^{\circ }$, since $AD$ is a diameter. We deduce that triangles $AGD$ and $ADB$ are similar, and therefore $\angle AFG=\angle CBA$.

Because $\angle AEB=\angle ADB=90^{\circ }$, quadrilateral $ABDE$ is cyclic. Therefore $\angle DEB=\angle CBA=\angle EFY$. But $\angle EFD=\angle YEF=90^{\circ }$. We deduce that $DFEY$ is a rectangle and therefore $BY$ is an altitude in triangle $BDZ$. Line segment $DG$ is also an altitude in triangle $BDZ$ which intersects $BY$ at $X$. We deduce that $ZX$ and $BC$ are perpendicular.



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Alternative solution.

Because $\angle ADB=\angle AEB=90^{\circ }$, the quadrilateral $ABDE$ is cyclic. Therefore, the projections of the point $D$ on the lines $AB,BE$, and $EA$ are collinear (Simson line). But $G$ and $F$ are the projections of $D$ on lines $AB$ and $EA$, respectively, since $AD$ is a diameter of the circle $\omega$. Then the projection of $D$ on $BE$ is $Y$, the intersection point of $BE$ with $GF$. Therefore, $BY$ is perpendicular to $DZ$. Hence, in triangle $BDZ$, line segments $BY$ and $DG$ are altitudes and intersect at $X$. Thus $ZX$ is also an altitude and therefore $ZX$ and $BC$ are perpendicular.