# In fig. 2, two ta

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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