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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.

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Solution

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 C1 and C2 with centre O, and AB is the chord of C1 touching C2 at C.
To prove : AC = CB
Construction : Join OC.
Proof : AB is the chord of C1 touching C2 at C, then AB is the tangent to C2 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).


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