Make a vertical line from p to ab, AC and BC. In d, f, h

△△DBP?△HBP, △ PFC △ PHC Because the distance between the points on the angular bisector is equal on both sides.

So DP=FP

In triangle ADP and triangle AFP, ∠ ADP = ∠ AFP = 90, DP=PF, AP=AP.

So triangle ADP≌ triangle AFP, so ∠BAP=∠CAP.

So p must be on the bisector of ∠ C.