Q. 35.0( 2 Votes )

# AP and AQ are two tangents drawn from an external point A to a circle with centre O, P and Q are points of contact. If PR is a diameter, let us prove that OA ∥ RQ

In Δ APO and Δ AQO

AP = AQ (Tangents drawn to the same circle from an external point are always equal in length)

AO = AO (Common)

OPA = OQA (Tangents are perpendicular to the line joining the centre)

So Δ APO and Δ AQO are congruent by S.A.S. axiom of congruency.

So from corresponding parts of congruent triangle we get

POA = QOA = x (Let)

POA + QOA = 2x

QOR = 1800-2x (Angle on a straight line)

OQR = ORQ (Δ OQR is isosceles)

OQR + ORQ = 1800-(1800-2x) = 2x

Since they are equal so

OQR = x = AOQ

OQR and AOQ forms a pair of alternate interior angles

So OA RQ

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