# In ∆ABC, the bisector of ∠A intersects BC in D.

Question:

In $\triangle A B C$, the bisector of $\angle A$ intersects $B C$ in $D$. If $A B=18 \mathrm{~cm}, A C=15 \mathrm{~cm}$ and $B C=22 \mathrm{~cm}$, find $B D$.

Solution:

Given $A B=18 \mathrm{~cm}, A C=15 \mathrm{~cm}$ and $B C=22 \mathrm{~cm}$.

In $\triangle A B C, A D$ the bisector of $\angle A$

$\frac{A B}{A C}=\frac{B D}{D C}$

$\frac{A B}{A C}=\frac{B D}{B C-B D}$

$\frac{18}{15}=\frac{B D}{22-B D}$

On cross multiplication, we get

$6(22-B D)=5 \times B D$

$132-6 B D=5 B D$

$132=5 B D+6 B D$

$132=11 B D$

BD = 12 cm

Hence, the value of BD is 12cm