Prove that $A, P, Q$ are collinear.

Question

Two circles $\omega_1$, $\omega_2$ intersect at $A, B$. An arbitrary line through $B$ meets $\omega_1$, $\omega_2$ at $C, D$ respectively. The points $E, F$ are chosen on $\omega_1$, $\omega_2$ respectively so that $CE = CB$, $BD = DF$. Suppose that $BF$ meets $\omega_1$ at $P$, and $BE$ meets $\omega_2$ at $Q$. Prove that $A, P , Q$ are collinear.

If you can show that $\angle QBP=\angle QAE$ Or $\angle QPB=\angle QEA$ then you would be done because $ABPE$ is a cyclic quad.

Hints, suggestions and solutions would be appreciated.

Taken from the $4$th Iranian Geometry olympiad

Answer 1

Let $\measuredangle CEB=\measuredangle EBC=\alpha$ and $\measuredangle PBE=\beta.$

Thus, $$\measuredangle PAB=180^{\circ}-\measuredangle PEB=\measuredangle EPP+\measuredangle PBE=2\alpha+\beta.$$ In another hand, $$\measuredangle BAQ=180^{\circ}-\measuredangle BFQ=180^{\circ}-(\measuredangle BFD+\measuredangle QFD)=$$ $$=180^{\circ}-(\measuredangle DBF+\measuredangle QBD)=180^{\circ}-(\alpha+\beta+\alpha)=180^{\circ}-(2\alpha+\beta)$$ and we are done!

Answer 2

All the red-marked angles are equal.

All the green-marked angles are equal. In particular, $\angle QAD = \angle CBQ$ because of exterior angle of a cyclic quadrilateral.

$\angle pink = \angle green - \angle QAB = \angle BAD = \angle red$. Result follows.