Indonesia
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). ^@
Answer:
- 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) ^@.
