# A circle touches the side BC of a triangle ABC at P and touches AB and AC when produced at Q and R respectively. Show that AQ = ½ (perimeter of triangle ABC).

Given: A circle touches the side BC of Δ ABC at P and AB, AC produced at Q and R respectively.
To Prove: AQ is half the perimeter of Δ ABC.
Proof:
Lengths of two tangents from an external points are equal.
∴ AQ = AR, BQ = BP and CP = CR
Perimeter of Δ ABC = AB + BC + AC
= AB + (BP + PC) + (AR - CR)
= (AB + QB) + PC + AQ - PC           (∵  AQ = AR, BQ = BP and CP = CR )
= AQ + AQ
= 2AQ
AQ is half the perimeter of Δ ABC.

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