Q. 303.9( 8 Votes )

Prove that the le

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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