62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour

Clearly, triangles $BEL$ and $CDL$ are isosceles, and angles $\angle CLD$ and $\angle BLE$ are vertical (fig. 3). Then $\angle BEL=\angle BLE=\angle CLD=\angle CDL$. Then $\angle AEB=\angle ALC$ as adjacent to equal angles. Also $\angle CAL=\angle BAL$, as $AL$ is a bisector.

Note that triangles $CLA$ and $BEA$ are similar by two angles. Then $\frac{AL}{AE}=\frac{AC}{AB}$, from where $AL=AE\cdot \frac{AC}{AB}$. Also note that triangles $BLA$ and $ACD$ are similar by two angles. Then $\frac{AL}{AD}=$ $$\frac{AB}{AC}$$