One part of the problem is very easy.

Note that $AD=\sqrt{p(p-a)}$, therefore: \[AE\cdot{AF}=AD^{2}\Longleftrightarrow \angle{ADE}=\angle{AFD}. \] It seems that we're done !

How are we done? I got that it remains to prove $AB=AC\leftrightarrow AD=AY$. But how?

dragonfire wrote: How are we done? I got that it remains to prove $AB=AC\leftrightarrow AD=AY$. But how? If $XY\parallel AB$ then $\angle{XDF}=\angle{YDA}=\angle{AFD}$ so $D$ should be the tangency point of the ($A$) ex-circle with $BC$. However, $AD\leq{m_{a}}$ thus $AB=AC$ (for (in any triangle) if $A^{\prime}$ is the point of tangency of the ex-circle with $BC$ then $M\in{[DA^{\prime}]}$, where $M$ is the midpoint of $BC$). The converse is quite obvious.

Sailor wrote: (for (in any triangle) if $A^{\prime}$ is the point of tangency of the ex-circle with $BC$ then $M\in{[DA^{\prime}]}$, where $M$ is the midpoint of $BC$). How do you prove this?

Remark. Generally, $pr^{2}+ar_{1}r_{2}=pr(r_{1}+r_{2})$ (see http://www.mathlinks.ro/Forum/viewtopic.php?p=637179#637279 ). If $r_{1}=r_{2}=\rho$, then $\rho=\frac{r}{a}\cdot\left[p-\sqrt{p(p-a)}\right]$. From the obvious relation $\rho=\frac{pr}{p+AD}$ obtain : $\frac{r}{a}\cdot\left[p-\sqrt{p(p-a)}\right]=\frac{pr}{p+AD}$ $\implies$ ${\frac{r}{a}\cdot\frac{ap}{p+\sqrt{p(p-a)}}=\frac{pr}{p+AD}}$ $\implies$ $p+\sqrt{p(p-a)}=p+AD$ $\implies$ $\boxed{AD=\sqrt{p(p-a)}}$.

can anyone post a complete solution please?