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.
∠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