<s>[INST] relation between inscribed and circumsdribed circle Let T be the triangle with side lengths $b_1,b_2,b_3$, and $r_{insc}$ and $r_{cir}$
be the radii of the inscribed and circumscribed circles,respectively,
I need to show that 
$$ \frac {r_{insc}}{r_{cir}}=\frac {(b_2+b_3-b_1)(b_3+b_1-b_2)(b_2+b_1-b_3)}{2b_1b_2b_3}$$
Maybe it is easy, but I forgot relations on triangles,
Please help me.
Thanks
 [/INST] Let's side of the triangle be $a,b,c$ and $R$ and $r$ are radii of circumscribed and inscribed circles, respectively.
The equaiton now looks like
$$\frac{r}{R} = \frac{(a+b-c)(a+c-b)(b+c-a)}{2abc}$$
We multiply both sides by $\frac{a+b+c}{8}$
$$\frac{r(a+b+c)}{8R} = \frac{a+b-c}{2} \times \frac{a+c-b}{2} \times \frac{b+c-a}{2} \times \frac{a+b+c}{2}\times \frac{1}{abc}$$
$$\frac{r(a+b+c)}{8R} = \frac{(s-c)(s-b)(s-a)s}{abc}$$
Where $s$ is the semiperimetar. By Heron's formula we know that
$$K = \sqrt{(s-c)(s-b)(s-a)s}$$
Where K is the are of the triangle.
$$\frac{r(a+b+c)}{8R} = \frac{K^2}{abc}$$
We substitute $\frac{a+b+c}{2} = s$
$$\frac{rs}{4R} = \frac{K^2}{abc}$$
We know that $K = \frac{abc}{4R} = rs$
$$\frac{rs \times abc}{4R} = K^2$$
$$K^2 = K^2$$
Q.E.D.

Another solution is making this substitution:
$$R = \frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$$
$$r = \frac{\sqrt{(s-a)(s-b)(s-c)}}{\sqrt{s}}$$
$$\frac{r}{R} = \frac{\frac{\sqrt{(s-a)(s-b)(s-c)}}{\sqrt{s}}}{\frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}} = \frac{4(s-a)(s-b)(s-c)}{abc} = \frac{4 \times \frac{c+b-a}{2} \times \frac{a+c-b}{2} \times \frac{a+b-c}{2}}{abc}$$
$$\frac{r}{R} = \frac{(a+b-c)(a+c-b)(c+b-a)}{2abc}$$
Q.E.D.
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