### A circle is touching the side $BC$ of $\triangle ABC$ at $P$ and touching $AB$ and $AC$ produced at $Q$ and $R$ respectively. Prove that $AQ = \dfrac { 1 } { 2 } \text{(Perimeter of } \triangle ABC)$. A B C P R Q

Thus, \begin{aligned} & AQ = AR && \ldots \text{(i)} && \text{[Tangents from A]} \\ & BP = BQ && \ldots \text{(ii)} && \text{[Tangents from B]} \\ & CP = CR && \ldots \text{(iii)} && \text{[Tangents from C]} \end{aligned}
2. We know that the perimeter of a triangle is the sum of the lengths of its sides. \begin{aligned} \text{ So, perimeter of } \triangle ABC & = AB + BC + AC \\ & = AB + (BP + CP) + AC && \text{[As, BC = BP + CP]} \\ & = AB + BQ + CR + AC && \text{[Using } eq \text{(ii) and } eq \text{(iii)]} \\ & = AQ + AR && \text{[As, AQ = AB + BQ and } \space \\ & && \text{AR = CR + AC] } \\ & = 2AQ && \text{[Using } eq \text{(i)]} \end{aligned}
3. Therefore, $AQ = \dfrac { 1 } { 2 } \text{(Perimeter of } \triangle ABC).$