In $$\triangle$$ABC, AD,the bisector of $$\angle$$A, meets BC at D. If BC = a, AC = b and AB =c, then BD - DC =
According to angle bisector theorem
$$\frac{AB}{AC}=\frac{BD}{DC}$$
$$\frac{c}{b}=\frac{BD}{DC}$$
Add 1 in both side
$$\frac{c}{b}+1=\frac{BD}{DC}+1$$
$$\frac{c+b}{b}=\frac{BD+DC}{DC}$$
$$\frac{c+b}{b}=\frac{a}{DC}$$
$$DC=\frac{ab}{b+c}$$ ------I
$$\frac{b}{c}=\frac{DC}{BC}$$
Add 1 in both side
$$\frac{b}{c}+1=\frac{DC}{DB}+1$$
$$\frac{c+b}{c}=\frac{DC+DB}{DB}$$
$$\frac{c+b}{c}=\frac{a}{DB}$$
$$DB=\frac{ac}{b+c}$$ ------ii
$$BD-DC=\frac{ac}{b+c}-\frac{ab}{b+c}=\frac{a\left(c-b\right)}{b+c}$$
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