Solutions of triangles - proof

Question

Question:

For a triangle ABC, prove that: $$r_1 + r_2 + r_3 = r + 4R$$ Where $r_1,r_2,r_3$ represent the radius of the ex-circles opposite to angle A, B, and C respectively. $r$ represents the radius of the incircle of the triangle and $R$ represents the radius of the circumcircle.

No idea where to even begin the question. I know that the: $$r_1 = \frac{\triangle}{S-a}$$

And similarly for the other radius of ex-circles. ($\triangle$ represents area of the triangle, $S$ represents the semi-perimeter of the triangle, and $a$ the side opposite to angle A)

Answer 1

This is my first answer in LaTeX. Please edit if wrong. So, rewrite the equation as $r_1-r+r_2+r_3=4R$. Substitute your formulates in the given equation and take $\Delta$ common and in the denominator, you should have $\Delta^2$ which means a net $\Delta$ in the denominator. In the numerator, again you should have an $a$ common and remaining inside the bracket should look like $s(s-a)+(s-b)(s-c)$. Open it up and you must have $2s^2-(a+b+c)s+bc$. The terms cancel out and you remain with $abc/\Delta$. And the result follows.

Answer 2

HINT:

Use

$$r=4R\sin\frac A2\sin\frac B2\sin\frac C2$$ and $$r_1=4R\sin\frac A2\cos\frac B2\cos\frac C2$$ etc.

Finally use $\cos(A+B),\sin(A+B)$ formulae