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 $60^\circ$.