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Answers
Let me first explain the proof strategy and then we'll prove it together. It is just another normal question, except that it expects you to make one interesting conclusion. Notice that OP passes through AB and O is the centre. Which theorem comes to your mind when you see that? "A perpendicular to a chord from the centre bisects the chord", right? A similar observation can be made for chord AC and line segment QP. But we can't directly apply that here, because we don't know if OP is perpendicular to AB or not. It's not given.
What we do know, however, is that AOPQ is a parallelogram. How can we use this fact to show that OP is perpendicular to AB? (Because if we can prove that, the rest is very simple). Now that I've made you think about the strategy, let me jump straight into the proof.
(Sorry if you expected a direct answer. The direct answer starts here. However, I highly recommend that you use the above hint and try solving it yourself before looking at the solution. At least try.)
Note that when I write PoC, I mean, Point of Contact.
Proof:
AB is a tangent to the circle with centre Q with A as the PoC [Given]
⇒ AQ ⊥ AB [Tangent is perpendicular to the radius at PoC] --(i)
Also, Since AOPQ is a parallelogram [Given],
⇒ AQ ║ OP [Opposite sides of a parallelogram are parallel.] --(ii)
Now, AB is the transversal between parallel lines AO and OP and AB ⊥ AQ [From (i)],
⇒ OP ⊥ AB {This was the interesting conclusion. Tadaa!} --(iii)
Now, AB is a chord and O is the center of the left circle. [Given] --(iv)
⇒ AP = BP [From (iii) and (iv); Perpendicular from the centre to a chord bisects the chord.]
i.e. OP is the perpendicular bisector of AB.
Similarly, without loss of generality, it can be proved that QP is the perpendicular bisector of AC.
This implies, P is the circumcenter of ΔABC. [Circumcenter is the point that passes through the perpendicular bisectors of the sides of a triangle]
Q.E.D.
I hope that helps :)