MathNet 2022-part-05 — 197 problems (EN–VI) Question 177

Let $AL$ be a bisector of triangle $ABC$. The circle centered at $B$ with radius $BL$ meets the ray $AL$ at point $E$, and the circle centered at $C$ with radius $CL$ meets the ray $AL$ at point $D$ (points $E$ and $D$ are different from point $L$). Prove that $AL^2 = AE \cdot AD$.
(Mykola Moroz)