Area of the triangle a,b,c How to prove in any triangle that the area $X$ is given by:

$$X=\frac{1}{4}(a+b+c)^2\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}$$
assume that c is the longest side and that d is perpendicular to c forming 2 right triangles.  The angle between a and d should be called D. The line along c from a to d is c'.

$$Area=\frac 14(a+b+c)^2\tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}$$
$$Area=\frac 14(a+c'+d)^2\tan\frac{D}{2}\tan\frac{B}{2}\tan45+\frac 14(b+d+c-c')^2\tan\frac{A}{2}\tan\frac{C-D}{2}\tan45$$
$$\tan 45=1$$
$$d^2=a^2-c'^2=b^2-(c-c')^2$$
IFF we can show the following that works, then the full one should too!
$$\frac 14(a+c'+d)^2\tan\frac{D}{2}\tan\frac{B}{2}=\frac 12c'd$$
$$B=90-D$$
$$\tan \frac B2 = \frac{\sin B}{1+\cos B}$$
$$\tan \frac D2 = \frac{\sin D}{1+\cos D}$$
$$\sin B = \cos D = \frac da$$
$$\sin D = \cos B = \frac {c'}a$$
$$\tan\frac{D}{2}\tan\frac{B}{2} = \frac{c'd}{a^2+ac'+ad+c'd}$$
$$\frac 14(a+c'+d)^2\frac{c'd}{a^2+ac'+ad+c'd}=\frac 12c'd$$
$$(a+c'+d)^2=2(a^2+ac'+ad+c'd)$$
$$a^2+c'^2+d^2+2ac'+2ad+2c'd=2a^2+2ac'+2ad+2c'd$$
$$a^2+c'^2+d^2=2a^2$$
$$c'^2+d^2=a^2$$
which is true.... if we repeat for the other right triangle this works.... Doing it from the Heron is easier but only, I guess, if you know the Heron and can prove it.This uses http://en.wikipedia.org/wiki/Tangent_half-angle_formula so .... yeah....
From this, $$\tan\frac A2=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\text{  where  } s=\frac{a+b+c}2$$

This can be easily derived using Tangent half-angle formula and the Law of cosines

Now we know Heron's formula