In the given figure, R is an external point and RP and RQ are tangents drawn to the circle from R. ∠POQ=90∘. The quadrilateral OPRQ is a
Square, 45∘
Given that ∠POQ=90∘
A tangent to the circle is perpendicular to the radius at the point of contact
Hence, ∠OPR=90∘ and ∠OQR=90∘
Now, in the quadrilateral OPRQ,
∠POQ+∠OQR+∠QRP+∠RPO=360∘.
90∘+90∘+∠QRP+90∘=360∘
Hence, ∠QRP=90∘.
Since, the angles are 90∘, OP is parallel to RQ and OQ is parallel to RP.
Hence, OPQR 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∘.
Hence, OPRQ is a square
Now, OR bisects ∠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, ∠ORP=∠ORQ=90∘2=90∘.
In the triangle OPR,
∠ORP+∠POR+∠OPR=180∘
⇒45∘+90∘+∠POR=180∘
Hence, ∠POR=45∘