Let us consider a circle with center O.

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

To Proof : PT = QT

Proof :

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.