If a circle touches the side $BC$ of a triangle $ABC$ at $P$ and extended sides $AB$ and $AC$ at $Q$ and $R$, respectively, prove that $AQ = \frac{1}{2}(BC + CA + AB)$.

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Question

If a circle touches the side $BC$ of a triangle $ABC$ at $P$ and extended sides $AB$ and $AC$ at $Q$ and $R$, respectively, prove that $AQ = \frac{1}{2}(BC + CA + AB)$.

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Accepted Answer

Step 1: $AQ = \frac{1}{2}(2AQ) = \frac{1}{2}(AQ + AQ)$
Step 2: $= \frac{1}{2}(AQ + AR)$ (since $AQ = AR$, tangents from external point)
Step 3: $= \frac{1}{2}(AB + BQ + AC + CR)$
Step 4: $= \frac{1}{2}(AB + BP + AC + CP)$ ($\because BQ = BP$, $CR = CP$, tangents from external points). $= \frac{1}{2}(AB + BC + CA)$

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