Hello everyone!

I have some serious problems with the next trigonometry problem.

I have to find pointPsuch that its pedal triangle is perspective to given triangleABC(see picture). If I understood correcty, triangles are prespective if extensions of their sides collide in three collinear points. I think that this pointPshould be in incenter of triangleABC. Now I have some trouble proving this statement.

I tried to start with two obvious facts:

$\displaystyle \angle C'DE + \angle BDC' + \angle BDF = 180^{\circ}$

$\displaystyle \angle BAP = \angle BC'P = \angle AP'B = 90^{\circ}$

Using these relations, I tried to prove that

$\displaystyle \angle A'C'B = \angle BA'C$

or $\displaystyle \angle AB'C' = \angle AC'B$

or $\displaystyle \angle A'B'C = \angle B'A'C$.

I'm trying to do it for a few last days but I just can't find the solution. I keep turning around the same thing again and again. I'm not even sure if my initial assumptions are enough to prove this or maybe I'm going in completely wrong direction.

I managed to prove this in opposite direction (if I assume that one of these three equations I wrote above are true) but it doesn't actually help a lot.

Please give me some clues, because I'm completely lost. Thanks in advance!