A chord PQ of a circle …

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Rahul Sharma 7 years, 10 months ago
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 ⊥<i>m </i>(Radius is perpendicular to the tangent at the point of contact)
PQ||<i>m</i> (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.
0Thank You