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eorem 9.2 The lengths of tangents drawn fromexternal point to a circle are equal.of: We are given a circle with centre O, a pointing outside the cirequired to prove that PQ PR.rcle and two tangents PQ, PR on the circle from

Answer»

Let two tangent PT and QT are drawn to circle of centre O as shown in figure. Both the given tangents PT and QT touch to the circle at P and Q respectively.

We have to proof : length of PT = length of QT Construction :- draw a line segment ,from centre O to external point T { touching point of two tangents } .

Now ∆POT and ∆QOT We know, tangent makes right angle with radius of circle. Here, PO and QO are radii . So, ∠OPT = ∠OQT = 90° Now, it is clear that both the triangles ∆POT and QOT are right angled triangle.nd a common hypotenuse OT of these [ as shown in figure ]

Now, come to the concept , ∆POT and ∆QOT ∠OPT = OQT = 90° Common hypotenuse OT And OP = OQ [ OP and OQ are radii]So, R - H - S rule of similarity ∆POT ~ ∆QOT Hence, OP/OQ = PT/QT = OT/OT PT/QT = 1 PT = QT [ hence proved]



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