Question

A circle is touching the side BC of $\Delta ABC$ at P and touching AB and AC produced at Q and R respectively. Prove that $AQ=\frac{1}{2}$ (perimeter of triangle ABC)

Answer 1

Given: A circle touching the side BC of $\Delta ABC$ at P and AB, AC produced at Q and R respectively.
To Prove: $AQ=\frac{1}{2}$ (Perimeter of $\Delta ABC$)
Proof: 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
= AB + (BP + PC) + (AR - CR)
= (AB + BQ) + (PC) + (AQ - PC) [AQ = AR, BQ = BP, CP = CR]
= AQ + AQ
= 2AQ
$\Rightarrow AQ=\frac{1}{2}$ (Perimeter of $\Delta ABC$)
$\therefore$ AQ is the half of the perimeter of $\Delta ABC$.