### In the given figure, the incircle of $\triangle ABC$ touches the sides $BC, CA,$ and $AB$ at $P, Q,$ and $R$ respectively. Prove that $(AR + BP + CQ) = (AQ + BR + CP) = \dfrac { 1 } { 2 } \text{(Perimeter of } \triangle ABC).$ R Q P B C A

Thus, \begin{aligned} & AR = AQ && \ldots \text{(i)} && \text{[Tangents from A] } \\ & BP = BR && \ldots \text{(ii)} && \text{[Tangents from B] } \\ & CQ = CP && \ldots \text{(iii)} && \text{[Tangents from C] } \end{aligned} Adding $\text{(i)}, \text{(ii)},$ and $\text{(iii)}$ equations, we have $$AR + BP + CQ = AQ + BR + CP = k \text{(say)}.$$
2. We know that the perimeter of a triangle is the sum of its sides. \begin{aligned} \text{ So, perimeter of } \triangle ABC & = AB + BC + CA \\ & = (AR + BR) + (BP + CP) + (CQ + AQ) \\ & = (AR + BP + CQ) + (AQ + BR + CP) \\ & = (k + k) = 2k \end{aligned} Thus, $k = \dfrac { 1 } { 2 } \text{ (Perimeter of } \triangle ABC)$
3. Thus, we can say that $(AR + BP + CQ) = (AQ + BR + CP) = \dfrac { 1 } { 2 } \text{ (Perimeter of } \triangle ABC)$.