Proving equalities with the incircle

Question

Prove that if $I$ is the incentre of the triangle $ABC$, then $$AI^2 = bc(s − a)/s.$$

My try:

$sr = (bc)(sinA)/2$ which is similar to the expression above, I equated both of the equalities,

$bc(sinA)/(sr)(2) = bc(s-a)/s(AI^2)$

we only have to prove

$a/s = (s-a)/AI^2$

which I need help proving

Answer 1

We have $${s-a\over AI} = \cos (A/2)$$ and $${r\over AI} = \sin (A/2)$$

so if we multiply those we get $${r(s-a)\over AI^2} = \sin (A/2)\cos (A/2) = \sin A/2$$

Now you can finish.

Answer 2

Draw the perpendicular fom I to $AC$ then we have

$$AI^2=r^2+AD^2$$ Now use that

$$r=\frac{A}{s}$$ and $$AD=s-a$$ where $$s=\frac{a+b+c}{2}$$