Question

The sides AB BC and CA of triangle ABC touch a circle with center O and radius r at P, Q and R respectively. Prove that area of triangle ABC = $\frac{r}{2}\times perimeteroftriangleABC$

Answer 1

We need to prove that
AB+CQ=AC+BQ
Since the sides of the triangles touches the circles,
they are the tangents of the circle.
Lengths of the tangents from the same external point are equal.
Thus, we have, BP=BQ, AP=AR and CQ=RC.
Now consider
AP+RC=RA+QC
Add BQ on both the sides of the equation, we have
$\Rightarrow AP+RC+BQ=RA+QC+BQ$
$\Rightarrow AP+QC+PB=RA+RC+BQ$ $[\because BQ=PB,QC=RC]$
$\Rightarrow AB+QC=AC+BQ$
Now,
$area\left(\Delta ABC\right)=area\left(\Delta OBC\right)+area\left(\Delta OCA\right)+area\left(\Delta AOB\right)$
$=\frac{1}{2}\times BC\times r+\frac{1}{2}\times CA\times r+\frac{1}{2}\times AB\times r$
$=\frac{r}{2}(AB+BC+CA)$