a circle touches the side BC of a triangle ABC at P and side AB and AC produced at Q and R respectively. prove that AQ= 1/2P(triangle ABC)

Asked by  | 30th Jan, 2012, 08:42: PM

Expert Answer:

The lengths of tangents drawn from an external point to a circle are equal. So, AQ = AR 2AQ= AQ +AR = (AB +BQ) + (AC + CR) = AB +AC +(BQ+CR) = AB +AC + (BP + CP) (As BQ = BP and CR = CP) So, 2AQ=AB+AC+BC 2AQ = Perimeter of triangle ABC Thus, AQ = 1/2(Perimeter of triangle ABC)

Answered by  | 31st Jan, 2012, 10:07: AM

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