Question

In fig. a circle touches the side BC of
triangle ABC at P and touches AB and
AC produced at Q and R respectively.
Show that AQ=1/2 (Perimeter of
∆ABC)​

Answer 1

Answer:

The lengths of the tangents,from a point to a circle,are equal.

Therefore AB = AP, BQ = BP, BP = CP

Perimeter = AC + AB + BC = 2(AB) + BP + PC = 2(AB) + 2(BQ) = 2(AQ) = 30 cm

thanks:)