Question

A circle touches the side $$BC$$ of $$\triangle\ ABC$$ at $$P$$ and touches $$AB$$ and $$AC$$ produced at $$Q$$ and $$R$$ respectively. Prove that $$AQ=\dfrac{1}{2}$$ (perimeter of $$\triangle\ ABC$$).

Solution

Given:A circle touching the side $$BC$$ of $$\triangle{ABC}$$ at $$P$$ and $$AB$$,$$AC$$ produced at $$Q$$ and $$R$$ respectively.Proof: Lengths of tangents drawn from an external point to a circle are equal.$$\Rightarrow AQ = AR, BQ = BP, CP = CR$$.Perimeter of $$\triangle{ABC}=AB + BC + CA$$$$= AB +\left(BP + PC\right)+\left(AR – CR\right)$$$$=\left(AB + BQ\right)+\left(PC\right)+\left(AQ – PC\right)$$ since $$\left[AQ = AR, BQ = BP, CP = CR\right]$$$$= AQ + AQ$$$$= 2AQ$$$$\Rightarrow AQ =\dfrac{1}{2}$$(Perimeter of $$\triangle{ABC}$$)$$\therefore AQ$$ is the half of the perimeter of $$\triangle{ABC}$$Mathematics

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