Q27 of 52 Page 10

In fig. 2, two tangents TP and TQ are drawn to a circle with centre O, from an external point T. Prove that (CBSE 2012)



Let ∠PTQ = x

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.


Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.


Property 3: Sum of all angles of a triangle = 180°.


By property 1,


TP = TQ (tangent from T)


Therefore, ∠TPQ = ∠TQP


Now,


By property 3 in ∆PAB,


∠TPQ + ∠TQP + ∠PTQ = 180°


⇒ ∠TPQ + ∠TQP = 180° – ∠PTQ


⇒ ∠TPQ + ∠TQP = 180° – x




By property 2,


∠TPO = 90°


Now,


∠ TPO = ∠ TPQ + ∠OPQ


⇒ ∠OPQ = ∠ TPO – ∠ TPQ





⇒ ∠PTQ = 2∠OPQ


Hence, Proved

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