Interesting proof of Heron's formula

Area of $\triangle ABC=\sqrt{s(s-a)(s-b)(s-c)}$, $s=\frac{1}{2}(a+b+c)$

$2s=2t+2(c-t)+2(b-t)$

$\begin{cases} s=b+c-t\:\:\:(1)\\2s=b+c+a\:\:\:(2)\end{cases}$

$(2)-(1), s=a+t \Rightarrow t=s-a$
$\text{From (1)}, s-b=c-t\:\text{and}\: s-c=b-t$

$\begin{cases}\tan \frac{A}{2}=\frac{R}{s-a} \\ \tan \frac{B}{2}=\frac{R}{s-b}\end{cases}\\ \Rightarrow \large \tan \frac{A}{2} \tan \frac{B}{2}=\frac{R^2}{(s-a)(s-b)}$

Similarly,
$\large{\tan \frac{A}{2} \tan \frac{C}{2}=\frac{R^2}{(s-a)(s-c)}\\ \tan \frac{B}{2} \tan \frac{C}{2}=\frac{R^2}{(s-b)(s-c)}}$

Adding all the three results,
$\large{\tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{A}{2} \tan \frac{C}{2}+\tan \frac{B}{2} \tan \frac{C}{2}\\ =R^2[\frac{1}{(s-a)(s-b)}+\frac{1}{(s-a)(s-c)}+\frac{1}{(s-b)(s-c)}]\\ =R^2[\frac{s-c+s-b+s-a}{(s-a)(s-b)(s-c)}]\\ =\frac{R^2s}{(s-a)(s-b)(s-c)}}$

Now, $A+B=\pi-C$.
$\frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}\\ \large{\tan (\frac{A}{2}+\frac{B}{2})=\frac{1}{\tan \frac{C}{2}}\\ \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}}\\ \tan \frac{A}{2} \tan \frac{C}{2}+\tan \frac{B}{2} \tan \frac{C}{2}=1-\tan \frac{A}{2} \tan \frac{B}{2}\\ \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{A}{2} \tan \frac{C}{2}+\tan \frac{B}{2} \tan \frac{C}{2}=1}$
$\large{\therefore 1=\frac{R^2s}{(s-a)(s-b)(s-c)}\\ R=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}}$

Area of $\triangle ABC=\frac{1}{2}(aR+bR+cR)=sR=\sqrt{s(s-a)(s-b)(s-c)}\:\Box$