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

Answer :


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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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