Question

In Fig.4, an isosceles triangle ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.

OR
In Fig.5, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

Answer 1

Given : An isoceles ABC with AB = AC, circumscribing a circle.
To prove : P bisects BC
Proof : AR and AQ are the tangents drawn from an external point A to the circle.
AR = AQ (Tangents drawn from an external point to the circle are equal)
Similarly, BR = BP and CP = CQ.
It is given that in ABC, AB = AC.
AR + RB = AQ + QC.
BR = QC (As AR = AQ)
BP = CP (As BR = BP and CP = CQ)
P bisects BC.
Hence, the result is proved.
OR

Given : Two concentric circles $C_1$ and $C_2$ with centre O, and AB is the chord of $C_1$ touching $C_2$ at C.
To prove : AC = CB
Construction : Join OC.
Proof : AB is the chord of $C_1$ touching $C_2$ at C, then AB is the tangent to $C_2$ at C with OC as its radius.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
OCAB
Considering, AB as the chord of the circle. So, OCAB.OC is the bisector of the chord AB.
Hence, AC = CB (Perpendicular from the centre to the chord bisects the chord).