How can I show PO is the angle bisector of XPY?

Question

A, B, X, and Y are points on circle. Q is the intersection point of AB and XY. PA and PB are tangents.

Answer 1

Extend PY to touch the circle at D. Mark intersection of PX and the circle as E. Connect D to E , this line intersect PO at C.We have :

$\angle XED=\angle XYD$

$\angle XCE=\angle DCY $

Therefore:

$\angle EDY=\angle YXE$

This is possible only if C is coincident with Q, hence points E and Y have equal distances from PO because PO is the symmetry axis of tangents PA and PB, therefore :

$\angle EQP=\angle YQP$

Hence triangles PQY and PQE are equal, so are the angles XPO and YPO, that means PO is the bisector of angle XPy.