Question

In $△$ ABC, ray AD bisect $\angle A$ and intersect BC in D. if BC= a, AC= b and AB=c, prove that
(i) $BD=\frac{ac}{b+c}$ (ii) $DC=\frac{ab}{b+c}$

Answer 1

In $△ABC$, $AD$ bisects $\angle A$

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

$\Rightarrow$ $\frac{b}{c}=\frac{DC}{BD}$

$\Rightarrow$ $\frac{b}{c}+1=\frac{DC}{BD}+1$ [ Adding 1 to both sides ]

$\Rightarrow$ $\frac{b+c}{c}=\frac{DC+BD}{BD}$

$\Rightarrow$ $\frac{b+c}{c}=\frac{BC}{BD}$ [ $DC+BD=BC$ ]

$\Rightarrow$ $\frac{b+c}{c}=\frac{a}{BD}$

$\therefore$ $BD=\frac{ac}{b+c}$ [ Hence proved ]

Similarly since $AD$ bisects $\angle A$.

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

$\Rightarrow$ $\frac{c}{b}=\frac{BD}{DC}$

$\Rightarrow$ $\frac{c}{b}+1=\frac{BD}{DC}+1$

$\Rightarrow$ $\frac{c+b}{b}=\frac{BD+DC}{DC}$

$\Rightarrow$ $\frac{c+b}{b}=\frac{BC}{DC}$

$\Rightarrow$ $\frac{c+b}{b}=\frac{a}{DC}$

$\therefore$ $DC=\frac{ab}{b+c}$ [ Hence proved ]