Question

A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisect the arc PRQ.

Answer 1

OR ⊥l (Radius is perpendicular to the tangent at the point of contact)

PQ||l     (given)

∴∠OSP = ∠OSQ = 90°  (pair of corresponding angles)

 no

In ΔOPS and ΔOQS

OP = OQ  (Radii of the same circle)

OS = OS  (Common)

∠OSP = ∠OSQ  (Proved)

So,ΔOPS ≅ ΔOQS  (RHS congruence criterion)

⇒ ∠POS = ∠QOS  (C.P.C.T)

⇒ arc (PR) = arc (QR)  (Measure of the arc is same as the angle subtended by the arc at the centre)
Thus, the point R bisects the arc (PRQ).