To two circles $r_1$ and $r_2$, intersecting at points $A$ and $B$, their common tangent $CD$ is drawn ($C$ and $D$ are tangency points, respectively, point $B$ is closer to line $CB$ than $A$). Line passing through $A$ , intersects $r_1$ and $r_2$ for second time at points $K$ and $L$, respectively ($A$ lies between $K$ and $L$). Lines $KC$ and $LD$ intersect at point $P$. Prove that $PB$ is the symmedian of triangle $KPL$. (Yu. Blinkov)