If the length of the sides BC, CA and AB of a △ABC are a, b and c respectively and AD

Let DC = x
∴ BD = a − x
By using angle bisector theore in △ABC, we have

$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}$

$\Rightarrow \frac{c}{b}=\frac{a-x}{x}$

$\Rightarrow c x=a b-b x$

$\Rightarrow x(b+c)=a b$

$\Rightarrow x=\frac{a b}{(b+c)}$

Now,

$a-x=a-\frac{a b}{b+c}$

$=\frac{a b+a c-a b}{b+c}$

$=\frac{a c}{(a+b)}$