Prove that sin A/2 * sin B/2 * sin C/2 = r/4R The other day I came across an identity in the book "Problems from the Book" and it was presented as well known:
$$\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} = \frac{r}{4R}$$
However I wasn't familiar with thee identity so I tried proving it.
Here is part of my attempt in solving this problem:
I first rewrote $\sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}$ as $\frac{r^3}{AI\cdot BI \cdot CI}$ (where I is the incenter of $\bigtriangleup ABC$). Then I tried using the fact that [ABC] =abc/4R (where [ABC] represents the area of $\bigtriangleup ABC$)which allowed me to rewrite the equation as $\frac{r^2}{AI\cdot BI \cdot CI} = \frac{[ABC]}{abc}$. I tried various things at this point but none of my attempts were successful...
Does anyone have a proof of this identity?

$$LHS = \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2} $$
$$= \sqrt{\frac{(s-b)(s-c)}{bc}} \cdot \sqrt{\frac{(s-c)(s-a)}{ca}}  \cdot \sqrt{\frac{(s-a)(s-b)}{ab}}$$
$$= \frac{(s-a)(s-b)(s-c)}{abc}$$
Note that $$r = \frac{\Delta}{s}$$
$$R = \frac{abc}{4 \Delta}$$
$$\Delta^2 = s(s-a)(s-b)(s-c)$$
Now $$ RHS = \frac{r}{4R} = \frac{\Delta}{s} \cdot \frac{\Delta}{abc} = \frac{(s-a)(s-b)(s-c)}{abc}$$
Hence LHS = RHS
