Given is triangle $ABC$ with $AB > AC$. Circles $O_B$, $O_C$ are inscribed in angle $BAC$ with $O_B$ tangent to $AB$ at $B$ and $O_C$ tangent to $AC$ at $C$. Tangent to $O_B$ from $C$ different than $AC$ intersects $AB$ at $K$ and tangent to $O_C$ from $B$ different than $AB$ intersects $AC$ at $L$. Line $KL$ and the angle bisector of $BAC$ intersect $BC$ at points $P$ and $M$, respectively. Prove that $BP = CM$.