Question 58
English source version: The circles $\omega_1$ and $\omega_2$ touch each other externally at $P$. Let $\ell_1$ be a line passing through the center of $\omega_1$ and tangent to $\omega_2$. Similarly, a line $\ell_2$ is tangent to $\omega_1$ and passes through the center of $\omega_2$. The lines $\ell_1$ and $\ell_2$ are not parallel. Prove that $P$ lies on the angle bisector of some angle formed by $\ell_1$ and $\ell_2$. (L. Emelyanov)
Russian source version, translated: Circles $\omega_1,\omega_2$ touch externally at $P$. Through the centre of $\omega_1$ draw a line $\ell_1$ tangent to $\omega_2$. Similarly, line $\ell_2$ is tangent to $\omega_1$ and passes through the centre of $\omega_2$. The lines $\ell_1,\ell_2$ turn out not to be parallel. Prove that $P$ lies on the bisector of one of the angles formed by these lines. (L. Emelyanov)
Russian source version, translated: Circles $\omega_1,\omega_2$ touch externally at $P$. Through the centre of $\omega_1$ draw a line $\ell_1$ tangent to $\omega_2$. Similarly, line $\ell_2$ is tangent to $\omega_1$ and passes through the centre of $\omega_2$. The lines $\ell_1,\ell_2$ turn out not to be parallel. Prove that $P$ lies on the bisector of one of the angles formed by these lines. (L. Emelyanov)