Let $AL$ be the inner bisector of triangle $ABC$. The circle centered at $B$ with radius $BL$ meets the ray $AL$ at points $L$ and $E$, and the circle centered at $C$ with radius $CL$ meets the ray $AL$ at points $L$ and $D$. Show that $AL^2 = AE\times AD$. (Proposed by Mykola Moroz)
