52
A circle touches the side BC of a $\triangle ABC$ at P and also touches the sides AB and AC produced at Q and R respectively. Prove that $AR = \frac{1}{2}$ (perimeter of $\triangle ABC$).
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Solution:
(a)
$AR = AQ$ (lengths of tangents from an external point are equal) (1/2 Mark)
$2AR = AQ + AR$
$= AB + BQ + AC + CR$ (1/2 Mark)
but $BQ = BP$
and $CR = CP$ } (lengths of tangents from an external point are equal ) (1/2 Mark)
$\therefore 2AR = AB + BP + AC + CP$
$= AB + BC+ AC = \text{Perimeter of } \triangle ABC$ (1 Mark)
$\therefore AR = \frac{1}{2}$ (Perimeter of $\triangle ABC$ ) (1/2 Mark)
(a)
$AR = AQ$ (lengths of tangents from an external point are equal) (1/2 Mark)
$2AR = AQ + AR$
$= AB + BQ + AC + CR$ (1/2 Mark)
but $BQ = BP$
and $CR = CP$ } (lengths of tangents from an external point are equal ) (1/2 Mark)
$\therefore 2AR = AB + BP + AC + CP$
$= AB + BC+ AC = \text{Perimeter of } \triangle ABC$ (1 Mark)
$\therefore AR = \frac{1}{2}$ (Perimeter of $\triangle ABC$ ) (1/2 Mark)