Question 61
In a triangle $ABC$, the interior angle bisector of $A$ intersects the $A$-excircle of $ABC$ at $D$ and $E$ such that $D \in [AE]$. Show that
$$
\frac{|AD|}{|AE|} \le \frac{|BC|^2}{|DE|^2}.
$$
$$
\frac{|AD|}{|AE|} \le \frac{|BC|^2}{|DE|^2}.
$$