Question

In the figure, $AB$ and $CD$ are two parallel tangents to a circle with centre $O$ and another tangent $PQ$ with point of contact $M$, intersecting $AB$ at $P$ and $CD$ at $Q$. Prove that $\angle POQ = 90^\circ$.

Accepted Answer

Join $OM$. In $	riangle OPA$ and $	riangle OPM$: - $AP = PM$ (tangents from P) - $OA = OM$ (radii) - $OP = OP$ (common) So $	riangle OPA \cong 	riangle OPM \implies \angle AOP = \angle MOP = 	heta_1$. Similarly, $	riangle OQC \cong 	riangle OQM \implies \angle COQ = \angle MOQ = 	heta_2$. Since $A-O-C$ is a straight line diameter (since $AB \parallel CD$): $$\angle AOP + \angle MOP + \angle MOQ + \angle COQ = 180^\circ$$ $$2	heta_1 + 2	heta_2 = 180^\circ \implies 	heta_1 + 	heta_2 = 90^\circ$$ Therefore, $\angle POQ = \angle MOP + \angle MOQ = 	heta_1 + 	heta_2 = 90^\circ$.

Angle subtended at center by parallel tangents segment

Explanation

Practise this under board exam conditions