Prove that the area of the triangle formed by the centers of the circumferences ex-inscribed is $\frac{abc}{2r}$

Question

Prove that the area of the triangle formed by the centers of the circumferences ex-inscribed is $\frac{abc}{2r}$, where $a,b,c,$ are the sides of the triangle and $r$ is the inradius.

I know that $S=s(r_a-a)\\ S=s(r_b-b)\\ S=s(r_c-c)\\ S=sr$

But I think this won't be useful... How can I prove this? Thanks for attention.

Answer 1

Before proving this, you need to prove few things (that I'm not proving) which will be helpful to prove the ultimate statement.

1. $\triangle ABC$ is the pedal triangle of $\triangle I_1I_2I_3$ with $I$ as it's orthocentre.

2. If $I_1, I_2$ and $I_3$ be the centres of the escribed circles which are opposite to the vertices $A, B$ and $C$ respectively of $\triangle ABC$. Then, $$I_1I_2=c\,\text{cosec}\frac{C}{2}$$

3. If $I$ is the incenter of $\triangle ABC$, then $$IA=\frac{r}{\sin\frac{A}{2}}$$

The area of $\triangle I_1I_2I_3$ is given as $A$

$\begin{align} &=\frac{1}{2}(IA) (I_2I_3)+\frac{1}{2}(IB) (I_3I_1)\frac{1}{2}(IC) (I_1I_2) \\ &=\frac{r}{2}\left(\frac{a}{\sin^2\frac{A}{2}}+\frac{b}{\sin^2\frac{B}{2}}+\frac{c}{\sin^2\frac{C}{2}}\right) \\ &=\frac{r}{2}\left(\frac{abc}{(s-b)(s-c)}+\frac{abc}{(s-a)(s-c)}+\frac{abc}{(s-a)(s-b)}\right) \\ &=\frac{abcr}{2}\left(\frac{s}{(s-a)(s-b)(s-c)}\right) \\ &=\frac{abcr}{2}\left(\frac{s^2}{\Delta^2}\right)\\ &=\frac{abc}{2r} \end{align}$

Here, $s=\displaystyle\frac{a+b+c}{2}$ and $\Delta=$ Area of $\triangle ABC$.

Answer 2

Let $\rho=\tfrac12(a+b+c)$, $a=|BC|$, $b=|AC|$, $c=|AB|$, $r$ and $r_a,r_b,r_c$ denote the inradius and the corresponding exradii of $\triangle ABC$,

\begin{align} r_a&=\frac{\rho r}{\rho-a} ,\quad r_b=\frac{\rho r}{\rho-b} ,\quad r_c=\frac{\rho r}{\rho-c} \tag{1}\label{1} . \end{align}

\begin{align} [O_aO_bO_c]&=[O_aCB]+[O_bAC]+[O_cBA]+[ABC] ,\\ &=\tfrac12(ar_a+br_b+cr_c)+\rho r \tag{2}\label{2} . \end{align}

Substitution of \eqref{1} into \eqref{2} gives

\begin{align} [O_aO_bO_c]&= \frac{\rho r\,(2\rho^3-(a+b+c)\rho^2+abc)}{2(\rho-a)(\rho-b)(\rho-c)} \tag{3}\label{3} ,\\ &= \frac{\rho^2 r\,(2\rho^3-2\rho\rho^2+abc)}{2\rho(\rho-a)(\rho-b)(\rho-c)} \tag{4}\label{4} ,\\ &= \frac{\rho^2 r\,abc}{2\rho^2r^2} = \frac{abc}{2r} \tag{5}\label{5} . \end{align}

The same area in terms of exradii is

\begin{align} [O_aO_bO_c]&= \frac{(r_a+r_b)(r_b+r_c)(r_c+r_a)}{2\sqrt{r_a r_b+r_b r_c+r_c r_a}} \tag{6}\label{6} . \end{align}