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