10. Let $A B$ be the diameter of a semicircle $\Gamma$. Two circles, $\omega_{1}$ and $\omega_{2}$, externally tangent to each other and internally tangent to $\Gamma$, are tangent to the line $A B$ at $P$ and $Q$, respectively, and to semicircular arc $A B$ at $C$ and $D$, respectively, with $A P<A Q$. Suppose $F$ lies on $\Gamma$ such that $\angle F Q B=\angle C Q A$ and that $\angle A B F=80^{\circ}$. Find $\angle P D Q$ in degrees.
Solution:
35
Extend the semicircle centered at $O$ to an entire circle $\omega$, and let the reflection of $F$ over $A B$ be $F^{\prime}$. Then $C Q F^{\prime}$ is a straight line. Also, the homothety centered at $C$ taking $\omega_{1}$ into $\omega$ takes $P$ to a point $X$ on $\omega$ and $A B$ to the parallel line tangent to $\omega$ at $X$. Therefore, $X$ is the midpoint of semicircle $A X B$, and $C, P$, and $X$ lie on a line. Similarly, $D, Q$, and $X$ lie on a line. So,
$$
45^{\circ}=\angle X C B=\angle P C B=\angle P C Q+\angle Q C B=\angle P C Q+10^{\circ},
$$
since $\angle Q C B=\angle F^{\prime} C B=\angle F^{\prime} A B=\angle F A B=90^{\circ}-\angle A B F=10^{\circ}$. Thus $\angle P C Q=35^{\circ}$. We will show that $\angle P C Q=\angle P D Q$ to get that $\angle P D Q=35^{\circ}$.
Note that $\angle X P Q$ subtends the sum of arcs $A C$ and $B X$, which is equal to arc $X C$. Therefore $\angle X P Q=\angle C D X$, so $C D Q P$ is cyclic and $\angle P C Q=\angle P D Q$. The conclusion follows.