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


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Answer:


Step by Step Explanation:
  1. We know that the lengths of tangents drawn from an external point to a circle are equal.
    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).^@

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