A circle touches the side $BC$ of a $ riangle ABC$ at a point $P$ and touches $AB$ and $AC$ when produced at $Q$ and $R$ respectively. Show that $AQ = \frac{1}{2}$(Perimeter of $ riangle ABC$).

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Question

A circle touches the side $BC$ of a $	riangle ABC$ at a point $P$ and touches $AB$ and $AC$ when produced at $Q$ and $R$ respectively. Show that $AQ = \frac{1}{2}$(Perimeter of $	riangle ABC$).

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Accepted Answer

Step 1: $AQ = AR$ (tangents from external point $A$ to the circle are equal).
Step 2: $2AQ = AQ + AR = (AB + BQ) + (AC + CR)$.
Step 3: $= AB + AC + BQ + CR = AB + AC + BP + CP$ (since $BQ = BP$ and $CR = CP$, tangents from $B$ and $C$).
Step 4: $= AB + AC + BC$. Therefore $AQ = \frac{1}{2}(AB + AC + BC) = \frac{1}{2}$(Perimeter of $	riangle ABC$).

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