$[ABC]=s \cdot r$ (where s is the semi-perimeter and r is the inradius)

Question

How do we prove $[ABC]=s \cdot r$ (where s is the semi-perimeter and r is the inradius). I refer to the area of ABC as [ABC].

Answer 1

Consider the diagram . We have :- $$[ABC] = [AIB]+[BIC]+[CIA] $$ Now , the red segments are perpendicular to the sides of the triangle . The segments of the same color are equal , as the tangents drawn from a point to a circle are equal .

Clearly , $$ [AIB]+[BIC]+[CIA] =( \frac{1}{2}AB \cdot IE)+ (\frac{1}{2} BC \cdot GI ) + (\frac{1}{2} CA \cdot FI ) = r\cdot\frac{1}{2}(AB+BC+CA)=\boxed{r\cdot s}$$

Answer 2

Suppose that $I$ is an incenter of triangle $ABC$. We will use the following formula for area: if $h$ is a distance from $X$ to line $YZ$ then area of $XYZ$ equals $\frac{YZ\cdot h}{2}$. Using this formula for triangles $AIB$, $BIC$ and $CIA$ (we consider $I$ as $X$) we get $$ [ABC]=[AIB]+[BIC]+[CIA]=\frac{r\cdot AB}{2}+\frac{r\cdot BC}{2}+\frac{r\cdot CA}{2}=\frac{r\cdot (AB+BC+CA)}{2}=r\cdot p, $$ as desired.