9. In triangle $A B C, \angle A B C=50^{\circ}$ and $\angle A C B=70^{\circ}$. Let $D$ be the midpoint of side $B C$. A circle is tangent to $B C$ at $B$ and is also tangent to segment $A D$; this circle instersects $A B$ again at $P$. Another circle is tangent to $B C$ at $C$ and is also tangent to segment $A D$; this circle intersects $A C$ again at $Q$. Find $\angle A P Q$ (in degrees).
Solution:
70

Suppose the circles are tangent to $A D$ at $E, F$, respectively; then, by equal tangents, $D E=D B=D C=D F \Rightarrow E=F$ (as shown). So, by the Power of a Point Theorem, $A P \cdot A B=A E^{2}=A F^{2}=A Q \cdot A C \Rightarrow A P / A Q=A C / A B \Rightarrow \triangle A P Q \sim \triangle A C B$, giving $\angle A P Q=\angle A C B=70^{\circ}$.