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 are PRQ.

Answer 1

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

To prove: The point R bisects the arc PRQ.

Construction: Join OR intersecting PQ at S.

Proof:

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

PQ || m (given)

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

In ΔOPS and ΔOQS

OP = OQ (Radius)

OS = OS (Common)

∠OSP = ∠OSQ

So,ΔOPS ≅ ΔOQS (RHS criterion)

⇒ ∠POS = ∠QOS (By C.P.C.T)

⇒ arc (PR) = arc (QR) (Angle subtended by the arc at the centre and measure of the arc is same)

∴ R bisect the arc PRQ.