Question

A circle is touching the side $BC$ of at $P$ and touching $AB$ & $AC$ produced at $Q$ and $R$ respectively. Prove that $AQ$ is the half of the perimeter of $\Delta ABC$.

Answer 1

Given that,

A circle touching the side $BC$ of $\Delta ABC$ at P and $AB$. $AC$ produced at $Q$ and $R$ respectively.

$AP=\frac{1}{2}$ (Perimeter of $\Delta ABC$)

So,

Lengths of tangents drawn from an external point to a circle are equal

$\Rightarrow AQ=AR,BQ=BP,CP=CR$

Perimeter of $\Delta ABC=AB+BC+CA$

$\begin{array}{c}\Delta ABC=AB+BC+CA\\ =AB+\left(BP+PC\right)+\left(AR-CR\right)\\ =\left(AB+BQ\right)+\left(PC\right)+\left(AQ-PC\right)\left[AQ=AR,BQ=BP,CP=CR\right]\\ =AQ+AQ\\ =2AQ\end{array}$

$\Rightarrow AQ=\frac{1}{2}$ (Perimeter of $\Delta ABC)$

$\therefore AQ$ is the half of the perimeter of $\Delta ABC$.

Hence, proved.