Answer :

Let us consider a circle with center O.

TP and TQ are two tangents from point T to the circle.

To Proof : PT = QT

OP ⏊ PT and OQ ⏊ QT [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

∠OPT = ∠OQT = 90°

In △TOP and △QOT

∠OPT = ∠OQT [Both 90°]

OP = OQ [Common]

OT = OT [Radii of same circle]

△TOP ≅ △QOT [By Right Angle-Hypotenuse-Side criterion]

PT = QT [Corresponding parts of congruent triangles are congruent]

Hence Proved.

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