Question

If r is the radius of incircle of $\Delta$ ABC, prove the following formula :
r = 4R sin(A/2) sin(B/2) sin(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, $ID=IE=IF=r$
$a=BC=BD+DC=r\cot \left(B/2\right)+r\cot \left(C/2\right)$
or 2R sin A
$=r\left(\frac{cos\left(B/2\right)\sin \left(C/2\right)+\cos \left(C/2\right)\sin \left(B/2\right)}{\sin \left(B/2\right)sin\left(C/2\right)}\right)$
or $2R.2\sin \left(A/2\right)\cos \left(A/2\right)\sin \left(B/2\right)\sin \left(C/2\right)=r.\sin (B+C)/2=r\cos \left(A/2\right)$
$\therefore 4Rsin\left(A/2\right)sin\left(B/2\right)sin\left(C/2\right)=r$,
$\because$ (B + C)/2 = ${90}^{\circ }-A/2$.