Question

An isosceles triangle ABC with AB=AC, circumcribes a circle. Prove that point of contact P bisects BC.

Answer 1

The tangents drawn from an externel point to a circle are equal in length.

∴ AP = AQ  ............ (1)

 BP = BR  ................ (2)

 CQ = CR    .............. (3)

Given that ABC is an isosceles triangle with AB = AC.

Subtract AP on both sides, we obtain

AB – AP = AC – AP

⇒ AB – AP = AC – AQ  (from (1))

∴ BP = CQ.

⇒ BR = CQ  (from (2))

⇒ BR = CR  (from (3))

 ∴BR = CR, it shows that BC is bisected at the point of contact R.

Answer 2

CRM

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