Let $D$ and $E$ be the midpoints of the sides $BC$ and $AC$ of a right triangle $ABC$. Prove that if $\angle CAD=\angle ABE$, then $$\frac{5}{6} \le \frac{AD}{AB}\le \frac{\sqrt{73}}{10}.$$
Problem
Source: 2006 All-Ukrainian Correspondence MO of magazine ''In the World of Mathematics'', grades 5-11 p7
Tags: geometry, geometric inequality, right triangle, Ukraine Correspondence