Internal bisector of an angle $\angle$A of triangle ABC meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E and the side AB at F. If a, b, c represent the sides of $∆$ABC then

By simple geometry, in $∆$AFE, we get AF = AE. Therefore, $∆$AFE, is an isosceles triangle. Now area 
$\left(\mathrm{△}\mathrm{ABC}\right)=\text{ area }\left(\mathrm{△}\mathrm{ABD}\right)+\text{ area }\left(\mathrm{△}ACD\right)$

$\Rightarrow \mathrm{AD}=\frac{2\mathrm{bccos}\frac{\mathrm{A}}{2}}{\mathrm{b}+\mathrm{c}}$
$\text{ Also, }\mathrm{AD}=\mathrm{AEcos}\frac{\mathrm{A}}{2}$
$\Rightarrow \mathrm{AE}=\frac{2\mathrm{bc}}{\mathrm{b}+\mathrm{c}}=\text{ H.M. of }\mathrm{b}\text{ and }\mathrm{c}$
$\text{ Again }\mathrm{EF}=2\mathrm{DE}=2\mathrm{AEsin}\frac{\mathrm{A}}{2}=\frac{4\mathrm{bcsin}\frac{\mathrm{A}}{2}}{\mathrm{b}+\mathrm{c}}$