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.

Answer 1

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

Now in right triangles $\Delta$ OQP and $\Delta$ORP,

OQ = OR (Radii of the same circle)

OP = OP (Common)

Therefore, $△$ OQP $\cong$ $△$ ORP (RHS)

This gives PQ = PR

We know that AP = AR

Given that AB = AC

i.e. AP + PB = AR + RC

$\Rightarrow$ 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

$\therefore$ BQ = CQ

The base is bisected by the point of contact