In the given figure, the incircle of ^@ \triangle ABC ^@ touches | Math Question and Answer

Answer:

Step by Step Explanation:

We know that the lengths of tangents from an external point to a circle are equal.
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 $AR + BP + CQ = AQ + BR + CP = k \text{(say)}.$
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, $k = \dfrac { 1 } { 2 } \text{ (Perimeter of } \triangle ABC)$
Thus, we can say that $(AR + BP + CQ) = (AQ + BR + CP) = \dfrac { 1 } { 2 } \text{ (Perimeter of } \triangle ABC)$ .

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