Two circles meet at $A$ (and $B$), and their common tangents meet at $O$. Show that $\overline{OA}$ bisects the angle made by the two tangents at $A$.

Question

Two circles are cut at $A$ and $B$, and their common tangents meet at $O$. If $AP$ and $AQ$ are the tangents at $A$ to the circles, how do you prove that $OA$ bisects $\angle PAQ$?

I have tried using the properties of circle that I knew of. I had a thought of orthogonal circles but couldn't proceed with it. Though I have been unable to bring coordinate geometry into picture, but I didn't get any clue of it.

Answer 1

Let $\Phi_1$ and $\Phi_2$ be our circles, $OA\cap\Phi_1=\{A,B\}$ and $OA\cap\Phi_2=\{A,C\}.$

We'll assume that $A$ placed between $B$ and $C$ and $B$ placed between $A$ and $O$.

Thus, $O$ is a center of a homotety $h$, for which $h(\Phi_1)=\Phi_2.$

Now, let $l_1$ and $l_2$ be tangents to $\Phi_1$ and $\Phi_2$ respectively such that $l_1\cap l_2=\{A\}$.

Since $h(A)=C$, we obtain $h(l_1)||l_1$ and $h(l_1)$ is a tangent line to $\Phi_2$.

Now, let $l_2\cap h(l_1)=\{D\}$.

Thus, $DC=DA$, $\measuredangle DCA=\measuredangle DAC$ and we are done!