Question

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

Answer 1

Given: Circle with centre O. PQ is the chord parallel to the tangent l at R

To prove: The point R bisects the arc PRQ.

Construction: Join OR intersecting PQ at S.

Proof:

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

PQ || l (given)

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

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).