# $(BDE) \cap (CDF) = P \not\equiv D$. 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.

• I have posted a proof that sadly doesn't incorporate your observation, it's really cool. How did you show that the points $B,E,H,K$ are concyclic? (I proved it by first proving $\angle AEF = \angle BKD$, but you seem to have done it the other way round, which is quite interesting). – Josef E. Greilhuber Sep 21 '19 at 20:16

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.