Selection tests for the Gulf Mathematical Olympiad, 2013

We have $\angle A F G=\angle A D G$ since $A G D F$ is cyclic. On the other hand $\angle D G A=\angle A F D=90^\circ$, since $A D$ is a diameter. We deduce that triangles $A G D$ and $A D B$ are similar, and therefore $\angle A F G=\angle C B A$.

Because $\angle A E B=\angle A D B=90^\circ$, quadrilateral $A B D E$ is cyclic. Therefore $\angle D E B=\angle C B A=\angle E F Y$. But $\angle E F D=\angle Y E F=90^\circ$. We deduce that $D F E Y$ is a rectangle and therefore $B Y$ is an altitude in triangle $B D Z$. Line segment $D G$ is also an altitude in triangle $B D Z$ which intersects $B Y$ at $X$. We deduce that $Z X$ and $B C$ are perpendicular.

Because $\angle A D B=\angle A E B=90^\circ$, the quadrilateral $A B D E$ is cyclic. Therefore, the projections of the point $D$ on the lines $A B, B E$, and $E A$ are collinear (Simson line). But $G$ and $F$ are the projections of $D$ on lines $A B$ and $E A$, respectively, since $A D$ is a diameter of the circle $\omega$. Then the projection of $D$ on $B E$ is $Y$, the intersection point of $B E$ with $G F$. Therefore, $B Y$ is perpendicular to $D Z$. Hence, in triangle $B D Z$, line segments $B Y$ and $D G$ are altitudes and intersect at $X$. Thus $Z X$ is also an altitude and therefore $Z X$ and $B C$ are perpendicular.