Team Selection Test

Since $C$, $D$, $P$, $F$ are cyclic, we have $\angle DPE=180^{\circ }-\angle C$.



As $A$, $E$, $P$, $F$ are cyclic, we also get $\angle FPE=180^{\circ }-\angle A$. Therefore, we have $\angle FPD=360^{\circ }-(180^{\circ }-\angle A+180^{\circ }-\angle C)=180^{\circ }-\angle B$ and hence, $B$, $D$, $P$, $F$ are cyclic as well.

Since $C$, $D$, $P$, $E$ are cyclic, we have $\frac{PE}{DC}=\frac{AE}{AD}$. Similarly, we have $\frac{PE}{AD}=\frac{AD}{AE}$. Since $B$, $D$, $P$, $F$ are cyclic. Thus, as $BD=DC$ we obtain $\frac{PE}{AD}=\frac{AD}{AE}$.

On the other hand, since $PQ$ is an angle bisector, we get $\frac{PE}{AE}=\frac{PF}{QE}$. $\frac{AF}{QE}=\frac{QF}{QE}$ and therefore, $\frac{AF}{AE}=\frac{QF}{QE}$ which implies that $AQ$ is the angle bisector of $A$ in triangle $EAF$.

Let $T$ be the intersection of $PQ$ and the line perpendicular to $BC$ passing through $A$. Since $AT$ is an altitude and $AQ$ is angle bisector in triangle $ABC$, we have $\angle TQA=(\angle B-\angle C)/2$. By angle chasing we have $\angle APE=\angle C$ and $\angle APF=\angle B$. As $PQ$ is an angle bisector in $FPE$, we get $\angle QPA=(\angle B-\angle C)/2$ and hence, we obtain $\angle TQA=\angle QPA$ which implies that $AT$ is tangent to the circumcircle of $AQP$ and we are done.