Points $A,B,C$ lie on a circle with center $O$, radius $7$. The perpendicular bisector of $AB$ meets $BC$ at $P$ and $AC$ at $Q$. Find $OP \cdot OQ$

Question

Points $A,B,C$ lie on a circle centered at $O$ with radius $7$. The perpendicular bisector of $AB$ meets $BC$ at $P$ and $AC$ at $Q$. Find $OP \cdot OQ$ .

What I Tried: Here is a picture :-

I thought about using Power of a Point but that did not help since $S$ does not lie on the circumference of the circle. The only information known is that the radius equals $7$, so I cannot use Pythagorean Theorem either. Joining $OA$,$OB$,$AP$ gives some congruent triangles, but I did not find a good way to use them.

Can anyone help me? Thank You.

Answer 1

Check that $\angle PCO = 90-A$. Also $\angle AQP = 90-A$.

Thus $OC$ is tangent to circumcircle of $\triangle PCQ$ (by converse of alternate segment theorem).

So by power of point, $$OP \cdot OQ = OC^2 = r^2$$

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Edit :

$\angle COA = 2B \Rightarrow \angle ACO = 90-B$

$\angle PCO = C - (90-B)=90-A$

Answer 2

Note $$\angle OCP = \frac{180 - \angle BOC }2 = 90 - \angle A = \angle Q$$

Thus, the triangles COP and COQ are similar, yielding

$$OP\cdot OQ = OC^2 =49$$