PB divides ∠ABC equally to get ∠ PBC = ∠ ABC/2;

PC divides ∠ACD equally to get ∠ PCD = ∠ ACD/2; Substitute (1) to get

∠ACD-∠ABC = 80；

In △ABC, ∠ BAC = ∠ ACD-∠ ABC = 80; (∠ACD is the outer corner); (2)

The perpendicular lines passing through point P, BC, AC and BA, intersect at D, E and F respectively;

∠ABC divided by PB gives PD = PF.

Divide ∠ACD by PC to get PE = PD.

It is deduced that PE=PF, so PA divides equally ∠ CAE; If the distance between a point and both sides of an angle is equal, then the line connecting it with the vertex of the angle is the bisector of the angle. )

That is ∠ cap = ∠ CAE/2; (3)

And ∠CAE is the outer corner of △ABC, where ∠ CAE =180-∠ BAC; Substitute (2) to get

∠ CAE = 100, substitute (3) to get:

∠CAP=50