Question

In the given figure, R is an external point and RP and RQ are tangents drawn to the circle from R. $\angle POQis{90}^{\circ }$. Prove that OPRQ is a prarallelogram, and find the $\angle$ POR.

Answer 1

Given that $\angle POQ$ = ${90}^{\circ }$
A tangent to the circle is perpendicular to the radius at the point of contact
Hence, $\angle OPR={90}^{\circ }and\angle OQR={90}^{\circ }$
Now, in the quadrilateral OPRQ,
$\angle POQ+\angle OQR+\angle QRP+\angle RPO={360}^{\circ }.$
${90}^{\circ }+{90}^{\circ }+\angle QRP+{90}^{\circ }={360}^{\circ }$
Hence, $\angle QRP={90}^{\circ }.$
Since the angles are ${90}^{\circ }$, OP is parallel to RQ and OQ is parallel to RP.
Hence, OPRQ is a parallelogram, and OQ = RP and OP = RQ;
Now, PR = RQ, since the lengths of tangents drawn from an external point to a circle are equal in length. And OP = OQ as they are radii of the same circle
Hence, OP = PR = RQ = QO. And all angles are ${90}^{\circ }$.
Hence, OPRQ is a square
Now, OR bisects $\angle$ PRQ since the line drawn from the center of the circle to an external point bisects the angle between the tangents drawn from that point.
Hence, $\angle ORP=\angle ORQ=\frac{{90}^{\circ }}{2}={45}^{\circ }.$
In the triangle OPR,
$\angle ORP+\angle POR+\angle OPR={180}^{\circ }$
${45}^{\circ }+{90}^{\circ }+\angle POR={180}^{\circ }$
Hence, $\angle POR={45}^{\circ }$