I am preparing for the BMO1 this year and found this geometry question 4 on the 1994 paper.
The points Q, R lie on the circle γ, and P is a point such that PQ, PR are tangents to γ. A is a point on the extension of PQ, and γ′ is the circumcircle of triangle PAR. The circle γ′ cuts γ again at B, and AR cuts γ at the point C. Prove that angle PAR = angle ABC.
I drew additional line segments PB and BR in γ′ to get the cyclic quadrilateral APBR and then drew BQ and BC in γ to get the cyclic quadrilateral QBRC. Then I did a lot of angle chasing using the tangent alternate segment circle theorem and found that triangle BPQ is similar to BQC and also triangle PAB is similar to BRC.
From here, I can't find a way to proceed. Please can someone give me a hint on what my next step should be (preferably a geometric solution).
Good luck for the BMO this year! This should be a matter of using the alternate segment theorem, the theorem that opposite angles in a cyclic quadrilateral are supplementary (twice), and that the triangle $PQR$ is isosceles because $PQ$ and $PR$ are tangents from the same external point, and chasing angles.
Diagram:
1: $\angle PAR = 180^{\circ} - \angle AQC - \angle QCA$ (angles sum to $180^{\circ}$ in a triangle)
2: $\angle QCA = 180^{\circ} - \angle QCR$ (angles on a straight line sum to $180^{\circ}$)
3: $\angle PAR = \angle QCR - \angle AQC$ (substituting line 2 into line 1)
4: $\angle AQC = 180^{\circ} - \angle RQC - \angle PQR$ (angles on a straight line sum to $180^{\circ}$)
5: $\angle PQR = \angle QCR$ (alternate segment theorem)
6: $\angle AQC = 180^{\circ} - \angle RQC - \angle QCR$ (substituting line 5 into line 4)
7: $\angle PAR = 2\angle QCR - 180^{\circ} + \angle RQC $ (substituting line 6 into line 3)
8: $\angle PAR = 2 \angle PQR - 180^{\circ} + \angle RQC$ (substituting line 5 into line 7)
9: $\angle QPR = 180^{\circ} - 2 \angle PQR$ (tangents from the same external point)
10: $\angle PAR = \angle RQC - \angle QPR$ (substituting line 9 into line 8)
11: $\angle RQC = 180^{\circ} - \angle CBR$ (opposite angles in the cyclic quadrilateral $QRBC$ add to $180^{\circ}$)
12: $\angle PAR = 180^{\circ} - \angle CBR - \angle QPR$ (substituting line 11 into line 10)
13: $\angle QPR = 180^{\circ} - (\angle CBR + \angle ABC) $ (opposite angles in the cyclic quadrilateral $PABR$ add to $180^{\circ}$)
14: $\angle PAR = \angle ABC$ (substituting line 13 into line 12)