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Question

Prove that the lengths of tangents drawn from an external point to a circle are equal. Using the above do the following:

ABC is an isosceles triangle in which AB = AC, circumscribed about a circle as shown in the fig. Prove that the base is bisected by the point of contact.


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Solution

OQP and ORP are right angles, because these are angles between the radii and tangents,

Now in right triangles Δ OQP and ΔORP,

OQ = OR (Radii of the same circle)

OP = OP (Common)

Therefore, OQP ORP (RHS)

This gives PQ = PR

We know that AP = AR

Given that AB = AC

i.e. AP + PB = AR + RC

PB = RC (since AP = AR)

As the lengths of tangents drawn from an external point to a circle are equal

PB = BQ and RC = CQ

BQ = CQ

The base is bisected by the point of contact


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