AoPS - Problem

$ P \equiv \omega_1 \cap \omega_2$ and $ M, N$ are the tangency points of $ \omega_1$ and $ \omega_2$ with a common external tangent. Inversion with center $ P$ and power $ PB \cdot PD$ takes $ (S_1)$ and the line $ l$ into themselves, the circles $ \omega_1$ and $ \omega_2$ go to two parallel lines $ k_1$ and $ k_2$ tangent to $ (S_1)$ and the circle $ (S_2)$ goes to a circle $ (S_2')$ tangent to $ k_1,k_2.$ Hence, $ (S_2)$ is congruent to its inverse $ (S_2').$ Further, $(S_2),(S_2')$ are symmetrical about $ P$ $\Longrightarrow$ $ PC \cdot PA = PB \cdot PD.$ By Casey's theorem for $ (\omega_1,\omega_2,D,B,S_1)$ and $ (\omega_1,\omega_2,A,C,S_2),$ we get $ DB = \frac {2PB \cdot PD}{MN} \ , \ AC = \frac {2PA \cdot PC}{MN}$ Since $ PC \cdot PA = PB \cdot PD$ $ \Longrightarrow$ $ AC = BD$ $ \Longrightarrow AB = CD.$