The medians of a right triangle $ABC$ ($\angle C = 90^{\circ}$) intersect at $M$. Point $L$ lies on the $AC$ such that $\angle ABL=\angle CBL$. It turned out that $\angle BML = 90^{\circ}$. Find the ration $AB : BC$.
Source: Belarusian National Olympiad 2021
Tags: ratio, geometry
The medians of a right triangle $ABC$ ($\angle C = 90^{\circ}$) intersect at $M$. Point $L$ lies on the $AC$ such that $\angle ABL=\angle CBL$. It turned out that $\angle BML = 90^{\circ}$. Find the ration $AB : BC$.