Question

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

Answer 1

In the figure shown,

Let the sides,

$AB=c$

$AC=b$

$BC=a$

$PC=x$

$PB=a-x$

Now, AQ and AR are two pair of tangents drawn from the same extrernal point A. Also, PC and CR are tangents from point C and PB and QB are tangents from point B.

Using the property that lenght of tangents drawn to a circle from a given external point is always the same, we can say that

$PC=CR=x$

$BP=QB=a-x$

$AQ=AR$

=>$AB+BQ=AC+CR$

=> $c+a-x=b+x$

=> $x=\frac{a-b+c}{2}$

Now,

$AQ=AR=b+x$

=> $b+\frac{a-b+c}{2}$

=> $\frac{a+b+c}{2}$

=> Semiperimeter of Triangle $ABC$.