I am given a triangle $\triangle ABC$ with side lengths $AB=c$, $AC=b$, $BC=a$.
$r$ is the radius of the incircle, and $r_a,r_b,r_c$ are the radii of the excircles.
The incircle divides the sides of the triangle to $a=y+z$, $b=x+z$, $c=x+y$.
$p$ is the semi-perimeter of $\triangle ABC$.
I need to prove $\frac{1}{r}=\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}$.
I managed to prove $\frac{2}{r}=\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}$ and I am pretty sure I did everything ok.
I used $S_{ABC}=r\cdot p=r_a \cdot x=r_b \cdot y=r_c \cdot z$.
Am I wrong? or is there something wrong about the problem?
$\endgroup$1 Answer
$\begingroup$Using the relations from your last line:
$$\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c} = \frac{x+y+z}{S_{ABC}}=\frac{p}{S_{ABC}}=\frac{1}{r}$$
$\endgroup$ 1
