(1) the intersection point p is the vertical line PF on the AC side, and the vertical foot is f.

Because PA is the ∠BAC bisector of pf⊥ac. pe⊥ab.

So PE=PF

Similarly, PF=PD.

So PE=PD

(2) Connect BP, because PD⊥BC, PE⊥AB.

So p is on the bisector of ∠ b.

(The equidistant points at both ends of an angle are on the bisector of this angle)

Satisfied with the teacher's answer, please accept it. Thank you.