In any triangle $\triangle ABC$, show that $4R\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})=r$ This is a problem problem I found in a JEE examination prep textbook, it was a "starred" question which I believe implies that it is more challenging than usual. It goes as follows:

In any triangle $\triangle ABC$, show that $$4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=r$$
Hint: $$2R^2\sin\left(A\right)\sin\left(B\right)\sin\left(C\right)=\Delta$$

Here is my attempt at it. I want to know if this is correct and if there any better alternative approaches to achieve the same result, please do share them!
We know that:
$$\Delta=rs$$
Using the given hint:
$$2R^2\sin\left(A\right)\sin\left(B\right)\sin\left(C\right)=rs$$
$$16R^2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=r\left(\frac{a+b+c}{2}\right)$$
$$16R\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=r\left(\sin\left(A\right)+\sin\left(B\right)+\sin\left(C\right)\right)$$
Now, focusing on the equation of the right hand side for a bit, we know that:
$$A+B+C=\pi$$
$$\frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$$
$$\sin\left(A\right)+\sin\left(B\right)+\sin\left(C\right)=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)+2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)$$
$$2\cos\left(\frac{C}{2}\right)\left(\cos\left(\frac{A-B}{2}\right)+\cos\left(\frac{A+B}{2}\right)\right)$$
$$4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)$$
Now substituting this back into the original problem:
$$16R\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\cos\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)=\left(4r\right)\left[\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)\right]$$
And that gives us:
$$4R\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)=r$$
 A: 
In the figure, $O$ is the circumcenter of $\Delta ABC$, $I$ is the in-center and $AI$ is extended to meet the circumcircle at $D$.
$\angle DIC= \frac{A}{2}+\frac{C}{2}=\frac{A+C}{2}=90^o-\frac{B}{2}$
$$\implies \frac{IC}{\sin B}=\frac{CD}{\sin \angle DIC}$$
$$\implies IC=\frac{CD \times \sin B}{\sin (90^o-\frac{B}{2})}$$
$$\implies IC=\frac{CD \times 2\sin \frac{B}{2} \cos \frac{B}{2}}{\cos\frac{B}{2}}$$
$$\implies IC=2CD \sin \frac{B}{2}    \tag{1}$$
Note that
$$\frac{CD}{\sin \frac{A}{2}} = 2R  $$
$$  \implies CD= 2R \sin \frac{A}{2}  \tag{2}$$
and
$$r=AE=IC \sin \frac{C}{2}  \tag{3}$$
$(1), (2), (3) \implies$
$$ r=4R \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$$
A: Your proof is fine.
Another approach, with a little less trigonometry, uses Heron's formula and the law of cosines.
Square both sides and use $\sin^2(x/2)=\frac{1-\cos(x)}{2}$ so you want:
$$2R^2(1-\cos A)(1-\cos B)(1-\cos C)=r^2$$
The law of cosines says:
$$\cos A=\frac{b^2+c^2-a^2}{2bc},$$ so $$1-\cos A=\frac{a^2-(b-c)^2}{2bc}=\frac{(a+b-c)(a+c-b)}{2bc}$$ and likewise for $B,C.$
Starting with the hint:
$$\Delta =2R^2\sin A\sin B\sin C.$$
Multiply by $4R\Delta$, and use the law of sines:
$$4R\Delta^2=\Delta abc$$
Use Heron's formula on the left and $\Delta=rs$ on the right to get:
$$4Rs\frac{(a+b-c)(a+c-b)(b+c-a)}{8}=rsabc.$$
Squaring and re-arranging, we get:
$$2R^2\frac{(a^2-(b-c)^2)(b^2-(a-c)^2)(c^2-(a-b)^2)}{8a^2b^2c^2}=r^2$$
Then, by the law of cosines result above, this can be rewritten as:
$$2R^2(1-\cos A)(1-\cos B)(1-\cos C)=r^2$$
Which is what we wanted.
A: Here is a rather simple geometrical approach that does not use the hint
Let the incenter of the triangle be $I$. Draw a perpenduclar line from $I$ to $BC$ and call the foot of the perpendicular $D$. Now, $IB$ is the angle bisector of $\angle CBA$ and $\triangle DBI$ is a right-angled triangle, thus,
$$ \cot\frac{B}{2} = \frac{BD}{ID} = \frac{BD}{r} \implies BD = r \cot\frac{B}{2}  $$
Similarly
$$ \cot\frac{C}{2} = \frac{CD}{ID} = \frac{CD}{r} \implies CD = r \cot\frac{C}{2}$$
Now
$$\begin{align}
BC &= BD + CD \\
\implies a &= r \left(\cot\frac{B}{2} + \cot\frac{C}{2}\right) \\
&= r \left( \frac{\cos\frac{B}{2}}{\sin\frac{B}{2}} + \frac{\cos\frac{C}{2}}{\sin\frac{C}{2}}\right) \\
&= r \left( \frac{\sin\frac{B+C}{2}}{\sin\frac{B}{2} \sin\frac{C}{2}} \right) \\
\implies 2R\sin A \left(\sin\frac{B}{2} \sin\frac{C}{2} \right) &= r \sin\frac{\pi - A}{2} \\
\implies 2R \left( 2\sin\frac{A}{2} \cos\frac{A}{2} \right) \sin\frac{B}{2} \sin\frac{C}{2} &= r \cos\frac{A}{2} \\
\therefore 4R \sin\frac{A}{2} \sin\frac{B}{2} \sin\frac{C}{2} &= r
\end{align}$$
Note that $\cos\dfrac{A}{2} \ne 0$ because $A \ne \pi$
A: By using the area formula
$$\Delta=\sqrt{s(s-a)bc}\sin{\frac{A}{2}},$$
at each vertex of the triangle, we can have
$$\Delta^6=s^3(s-a)(s-b)(s-c)(abc)^2\sin^2{\frac{A}{2}}\sin^2{\frac{B}{2}}\sin^2{\frac{C}{2}}.$$
After simplifying by Heron's formula,
$$\Delta^4=s^2(abc)^2\sin^2{\frac{A}{2}}\sin^2{\frac{B}{2}}\sin^2{\frac{C}{2}}.\tag{1}$$
On the other hand,
$$\Delta^4=\Delta^2\Delta^2=(sr)^2\left(\frac{abc}{4R}\right)^2.\tag{2}$$
From $(1)$ and $(2)$, the result follows.
