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

                            

                                            OR

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

                                   

Given: ABC is an isosceles triangle, where AB = AC, circumscribing a circle. 

To prove: The point of contact P bisects the base BC, i.e. BP = PC

Proof: It can be observe that

BP and BR;  CP and CQ;  AR and AQ; are pairs of tangents drawn to the circle

from the external points B,  C, and A respectively.

Since the tangents drawn from an external point to a circle, then

                          BP = BR        ............(i)

                          CP = CQ        ............(ii)

                          AR = AQ        ............(iii)

Given that  AB = AC

$\Rightarrow$ AR + BR = AQ + CQ

$\Rightarrow$ BR = CQ             .........[ From (iii) ]

$\Rightarrow$ BP = CP              .........[From (i) and (ii) ]

$\therefore$  P bisects BC.

                                     OR  

Given: The chord AB of the larger of the two concentric circles, with centre O,

touches the smaller circle ar C.

To prove: AC = CB

Construction: Let us join OC.

Proof: In the smaller circle, AB is a tangent to the circle at the point of contact C.

$\therefore$ OC $\perp$ AB        .....................(i)

( Using the property that the radius of a circle is perpendicular to the tangent at the point of contact )

For the larger circle, AB is a chord and from (i) we have OC$\perp$AB 

$\therefore$ OC bisects AB

( Using the property that the perpendicular drawn from the centre to a chord  of  a circle bisects the chord.)

$\therefore$ AC = CB.