(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.