A question on geometry Two circles $c_1$ and $c_2$ intersect each other at points $A$ and $B$. Their external common tangent closer to $B$ touches $c_1$ at $P$ and $c_2$ at $Q$. Let point $C$ be the reflection of $B$ in the line $PQ$. Prove that angle $CAP$ = angle $BAQ$.
 A: Assume circle $c_1$ has smaller radius than $c_2.$ Extend the two common tangents to $c_1$ and $c_2$ until they intersect at a common point $O$ and draw the circle $c_O$ with center $O$ passing thorough point $A$. Then $c_O$ also passes through $B$ and $C$. consequently, $$\angle \, BAC = \frac{1}{2} \, \angle \, BOC = \angle \, BOP$$ Let $\angle \, BAP = \alpha$ and $\angle \, BAQ = \beta$. Then $\angle \, BPQ = \angle \, BAP = \alpha$ and $\angle \, BQP = \angle \, BAQ = \beta$ because these are angles associated to the common tangent $PQ$ to $c_1$ and $c_2$.
Let $OB$ intersect $c_1$ and $c_2$ in $B_1$ and $B_2$, respectively, as second points of intersection. If you perform a homothety with center $O$ mapping $c_1$ to $c_2$ then $B_1$ is mapped to $B$ and $B$ is mapped to $B_2$, while $P$ is mapped to $Q$ (tangent points). Therefore 
$$\angle \, OBP = \angle \, B_1BP = \angle \, BB_2Q = \angle \, BQP = \beta$$ 
Form triangle $OBP$
$$\angle \, BOP = \angle \, BPQ - \angle \, OBP = \alpha - \beta$$
Then
$$\angle \, BAC = \angle \, BOP = \alpha - \beta$$
Consequently, $$\angle \, CAP = \angle \, BAP - \angle \, BAC = \alpha - (\alpha - \beta) = \beta = \angle \, BAQ$$

Alternatively, if you are a fan of inversions, you can perform inversion with respect to $c_O$. Then circle $c_1$ is mapped to circle $c_2$ (and vice versa) and point $P$ is mapped to $Q$. Therefore if you wish $OP \cdot OQ = OB^2$ which leads to $\angle \, OBP = \angle \, BQP = \beta$ or if you prefer the circle $c^*$ passing through $P, B, Q$ is mapped to itself, so it is orthogonal to $c_O$ which means that $OB$ is tangent to circle $c^*$ at point $B$ and again  $\angle \, OBP = \angle \, BQP = \beta$.
