Answer :

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 = 2OPQ


Hence, Proved


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