Three Lines, Three Circles

@Calvin Lin @Michael Enright : Here it is.

Upgrade to IE 1 to view this content.

First, we will prove that

$[ABC] = r_a (s - a),$

where $[ABC]$ is the area of $ABC$ , $r_a$ is the radius of the A-excircle, and $s$ is the semiperimeter. (Skip to the black square if you want to get to the main proof.)

Let $X$ be the tangent point of the A-excircle with $BC$ , as shown above. Also, let $B_1$ be the tangent to $AB$ and $C_1$ be the tangent to $AC$ .

We will find the area of $ABC$ as the area of right triangles $AB_1I_A$ and $AC_1I_A$ minus the area of $B_1BCC_1I_A$ .

First, $BX = s - c$ and $CX = s - b$ , as shown in this wiki . Because tangents from a point to a circle are equal, we also have $BB_1 = s - c$ and $CC_1 = s - b$ . Thus, $AB_1 = s = AC_1$ and

$\begin{aligned} [AB_1I_A] + [AC_1I_A] &= \frac{(AB_1)(B_1I_A) + (AC_1)(C_1I_A)}{2} \\ &= \frac{sr_a + sr_a}{2} \\ &= sr_a . \end{aligned}$

To find $[B_1BCC_1I_A]$ , we note that $\triangle BXI_A \cong \triangle BB_1I_A$ and $\triangle CXI_A \cong \triangle CC_1I_A$ . So,

$\begin{aligned} [B_1BCC_1I_A] &= [BXI_A] + [BB_1I_A] + [CXI_A] + [CC_1I_A] \\ &= 2([BXI_A] + [CXI_A]) \\ &= 2[BCI_A] \\ &= ar_a. \end{aligned}$

Finally,

$\begin{aligned} [ABC] &= [AB_1I_A] + [AC_1I_A] - [B_1BCC_1I_A] \\ &= sr_a - ar_a \\ &= r_a(s - a), \end{aligned}$

as desired. $\blacksquare$

We have $r_a = \frac{[ABC]}{s - a}$ . Similarly, $r_b = \frac{[ABC]}{s - b}$ and $r_c = \frac{[ABC]}{s - c}$ . Also, $[ABC] = \sqrt{s(s-a)(s-b)(s-c)}$ .

Thus, we have

$\begin{aligned} r_ar_b + r_br_c + r_cr_a &= \frac{[ABC]^2}{(s-a)(s-b)} + \frac{[ABC]^2}{(s-b)(s-c)} + \frac{[ABC]^2}{(s-c)(s-a)} \\ &= \frac{((s-c) + (s-a) + (s-b))[ABC]^2}{(s-a)(s-b)(s-c)} \\ &= \frac{(3s - 2s)(s(s-a)(s-b)(s-c))}{(s-a)(s-b)(s-c)} \\ &= s^2 . \end{aligned}$