Prove that $\angle AQP = \angle ABC$ Given a triangle $ABC$ with $O$ as its circumcenter. Points $P$ and $C$ are the intersection points of the circumcircle of triangle $BOC$ and the circle with diameter $AC$. Point $Q$ lies on segment $PC$ such that $PB=PQ$. Prove that $\angle AQP = \angle ABC$.
Well, I tried this for a long time, and I am stuck till here :
Let $D$ be the foot of the perpendicular from $A$ to $BC$, since we need to prove that $\angle AQP = \angle ABC$, then we shall prove that $\angle PAQ = \angle BAH$, but I don't know what to do next, I have tried like some angke chasing but can't derive anything from the fact that $PB = PQ$, can anyone help? It will be really appreciated. Thanks a lot!!
 A:  
Hint. If triangle is isosceles, as can be seen in my drawing , right triangles APQ and BPC are equal because $BP= PQ$ and $PC=PA$, therefore $\angle PBC=\angle PQA$. For other cases we can do following methods:
1-Extend AQ to cross BC at D, also extend PC to intersect AB at K. Now if $\angle QDC=\angle BKQ$ then $\angle QDB + \angle BKQ=180^o$,  So $\angle KQD+\angle KBD=180^o$ that results in $\angle KBC=\angle PQA$.
2-Take point k on AB such that $BK=BP$. Draw a perpendicular from K on AB, it intersect BC at F. If we show $PA=KF$ then right triangles PAQ and KBF are equal then $\angle ABC=\angle PQA$.
A: We will denote by $x,y$ the (measures of the) angles in $A$, respectively $B$.

Let $R$ be the intersection of the line $CQP$ with the circle $(ABC)$.

 
Then we finish in few lines.


*

*The angle at center $\widehat {BOC}$ is twice the "inscribed" angle $A$, so
$$\widehat{BPC}=\widehat {BOC}=2x\ .$$

*The angles in $R,A$ that cover the chord $BC$ are equal, so both are $x$.

*The $2x$ angle in $P$ is the sum of the angles in $B,R$ in $\Delta PBR$, so the angle in $B$ in this triangle is $x$.

*The triangle $\Delta PBR$ is isosceles, so we get 
$$ PR = PB = PQ\ . $$

*The triangles $\Delta APR$ and $\Delta APQ$ are thus equal, so the angles in $R,Q$ are equal. This gives
$$
\widehat{AQP} =
\widehat{ARP} =
\widehat{ARC} =
\widehat{ABC} \ .
$$
$\square$
A: Violet circle (center V, diameter AC, through P).
Orange circle (center O, . , through A, B, C).
Green circle (center G, diameter OZ, through O, P, B, C) as shown.

A[V]C, P[M]C, B[N]C are respective common chords. OV, V[M]G, O[N]GZ are the lines of centers. ONG is extended to cut the green circle at Z. 
Extend CP to cut circle V at P’. Then, $\angle 1 = \angle 2$.
$\angle ABC = x = y = z$. This means BA // NV. Then, $\angle 2 = \angle 3 = \angle 4 = \angle 5$. This further means PZ // P’B.
Note also that $\omega = \theta = \phi$. Since $\phi = \angle 4$, we have $\angle 1 = \omega$. That is, PP’ = PB = PQ. Finally, $\angle ABC = x = t$.
