It is proved that the passing point P is PE ⊥ AM, PF ⊥ An, PO ⊥ BC, and the intersection points are M, N and O respectively.

Because BP is the bisector of the outer angle of △ABC, PM=PO.

And because CP is the bisector of the external angle of △ABC, PN=PO.

So PM=PN

So p must be on the bisector of ∠BAC.