$(BDE) \cap (CDF) = P \not\equiv D$. Prove that $PD$ passes through the excenter of $\triangle ABC$ in $\angle A$. 
$D$, $E$ and $F$ are points respectively on sides $BC$, $AB$ and $AC$ such that $AE = CD$ and $AF = BD$. $(BDE) \cap (CDF) = P \not\equiv D$. Prove that $PD$ passes through the excenter of $\triangle ABC$ in $\angle A$.


I have noticed that if $K$ is the midpoint of arc $BC$ not containing point $A$ and $EF \cap KD = H$ then $B, E, H, K$ and $C, F, H, K$ are concyclic.
$\implies \angle AEF = \angle BKD$ and $\angle AFE = \angle CKD$.
Not sure what is next though.

 A: The excenter is the radical center of the circumcircle $(ABC)$ and the two circles $(BDE)$ and $(CDF)$, thus the radical axis $DP$ of the latter two circles has to pass through it.
To prove this, let $N$ be the midpoint of the arc $AB$ of the circumcircle which contains $C$. Then $NA = NB$ holds, and by the inscribed angle theorem, $\angle DBN = \angle FAN$. Together with $DB = FA$, this implies that the triangles $DBN$ and $FAN$ are congruent (and have the same orientation), and hence $\angle BND = \angle ANF$.
This yields, together with the inscribed angle theorem,
$$\angle FCD = \angle ACD = \angle AND = \angle ANF + \angle FND - \angle BND = \angle FND,$$
and thus $N$ lies on the circle $(CDF)$. Therefore, the radical axis of $(ABC)$ and $(CDF)$ is the line $CN$, which is the exterior angle bisector in $C$ (this is well known, but also immediate from the inscribed angle theorem, since $AN=BN$). Analogously, the radical axis of $(ABC)$ and $(BDE)$ is the exterior angle bisector in $B$, and thus the radical center of the three circles is the excenter, as claimed.

