Question

In $△ABC$. $\overrightarrow{AD}$ the bisector of $\angle A$ intersects $\overline{BC}$ in $D$. $\therefore BD=..........$

• A. $\frac{BC\times AB}{AB+AC}$
• B. $\frac{BC\times AB}{AB-AC}$
• C. $\frac{BC\times AB}{AC-AB}$
• D. $\frac{BC\times AB}{BC+AB}$

Answer 1

Answer: (A)

$\frac{BC\times AB}{AB+AC}$

In $△ABC,$

$AD$ bisects $\angle A$

$\therefore$ $\frac{AC}{AB}=\frac{DC}{BD}$ [ By angle bisector theorem ]

$\Rightarrow$ $1+\frac{AC}{AB}=1+\frac{DC}{BD}$

$\Rightarrow$ $\frac{AB+AC}{AB}=\frac{BD+DC}{BD}$

$\Rightarrow$ $\frac{AB+AC}{AB}=\frac{BC}{BD}$ [ Since, $BD+DC=BC$ ]

$\Rightarrow$ $BD=\frac{AB\times BC}{AB+AC}$