Prove $\frac{r_1}{bc}+\frac{r_2}{ac}+\frac{r_3}{ab}=\frac1r-\frac1{2R}$ Prove $\dfrac{r_1}{bc}+\dfrac{r_2}{ac}+\dfrac{r_3}{ab}=\dfrac{1}{r}-\dfrac{1}{2R}$ where $a , b, c$ are sides of a triangle; $r $is the inradius; $R$ is the circumradius; the $r_i$ are the exradii; $s$ is semiperimeter; and $\triangle$ is area
Solving L.H.S
$$\dfrac{1}{abc}\left(ar_1+br_2+cr_3\right)$$
$$\dfrac{\triangle}{abc}\left(\dfrac{a}{s-a}+\dfrac{b}{s-b}+\dfrac{c}{s-c}\right)$$
$$\dfrac{\triangle}{abc}\left(\dfrac{a-s+s}{s-a}+\dfrac{b-s+s}{s-b}+\dfrac{c-s+s}{s-c}\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{1}{s-a}+\dfrac{1}{s-b}+\dfrac{1}{s-c}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{2s-a-b}{(s-a)(s-b)}+\dfrac{1}{s-c}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{c}{(s-a)(s-b)}+\dfrac{1}{s-c}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{c(s-c)+(s-a)(s-b)}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{cs-c^2+s^2-bs-as+ab}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{(s-c)(s+c)-s(a+b-c)+ab}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{(s-c)(s+c)-2s(s-c)+ab}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{(s-c)(c-s)+ab}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{2cs-s^2-c^2+ab}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{c(a+b+c)-s^2-c^2+ab}{(s-a)(s-b)(s-c)}\right)\right)$$
$$\dfrac{\triangle}{abc}\left(-3+s\left(\dfrac{ab+bc+ca-s^2}{(s-a)(s-b)(s-c)}\right)\right)$$
Now how to proceed from here, I just want the way to continue from here. Please don't share other ways, I particularly want to know the way to continue from the last step as in multiple questions I get to such points and don't find the way to proceed from there. 

As $r_1=s\tan\dfrac A2$
$T(1,a)=\dfrac{r_1}{bc}=\dfrac{2Rs\sin A\tan\dfrac A2}{abc}=\dfrac{2Rs(1-\cos A)}{abc}=\dfrac{2R\triangle(1-\cos A)}{r\cdot4R\triangle}=\dfrac{1-\cos A}{2r}$
using $\triangle=rs=\dfrac{abc}{4R}$
$T(1,a)+T(2,b)+T(3,c) =\dfrac{3-(\cos A+\cos B+\cos C)}r$
Now use Prove trigonometry identity for $\cos A+\cos B+\cos C$
and from here, $r=4R\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2\implies\cos A+\cos B+\cos C=1+\dfrac r{4R}$
