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]
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