Question

If r is the radius of incircle of $\Delta$ ABC, prove the following formula :
r = (s - a) tan(A/2) = (s - b) tan(B/2) = (s - c) tan(C/2)

Answer 1

The bisectors of the angles of $\Delta$ ABC meet in I. Draw ID, IE, IF perpendicular to the sides from I.
By Geometry, ID = IE = IF = r
We know by geometry that
BD = BF = $\beta$ (say), CD = CF = $\gamma$, AE = AF = $\alpha$
$\therefore 2\alpha +2\beta +2\gamma$ = a + b + c = 2s
or $\alpha +\beta +\gamma =s,$
$\therefore$ AF = $\alpha$ = s-($\beta +\gamma$) = s - BC = s - a
or BD = $\beta$ = s - ($\gamma +\alpha$) = s - AC = s - b,
CD = $\gamma$ = s -($\alpha -\beta$) = s - AB = s - c.
Now $\frac{ID}{BD}=tan\frac{B}{2},$
$\therefore r=\beta tan\frac{B}{2}=(s-b)tan\frac{b}{2}$
Similarly, $\frac{ID}{CD}=tan\frac{C}{2}$
$\therefore r=\gamma tan\frac{C}{2}=(s-c)tan\frac{C}{2}and\frac{IF}{AF}=tan\frac{A}{2}$,
$\therefore r=\alpha tan\frac{A}{2}=(s-a)tan\frac{A}{2}$