66th Czech and Slovak Mathematical Olympiad

We denote $\angle BAC$ by $\alpha$ and express lengths $AE$, $AD$, $BE$, $CD$ in terms of $\cos \alpha$ and the side lengths $b=AC$, $c=AB$ of triangle $ABC$. The condition rewrites as

$$b\cos \alpha \cdot c\cos \alpha =(c-b\cos \alpha )(b-c\cos \alpha ),$$ which simplifies to $bc=(b^2+c^2)\cos \alpha$. Hence $$\cos \alpha =\frac{bc}{b^2+c^2}\le \frac{1}{2},$$ where the last inequality is for any positive $b$, $c$ equivalent with an obvious inequality $(b-c{)}^2\ge 0$ (alternatively, one can use AM-GM inequality for $b^2$ and $c^2$). We proved that angle $BAC$ of any such triangle $ABC$ satisfies $\cos \angle BAC\le 1/2$, therefore $\angle BAC\ge 60^{\circ }$. Since for equilateral triangle, the condition is clearly satisfied (in that case $AE=AD=BE=CD$), the answer is $60^{\circ }$.