In Fig. 4.60, AD bisects ∠A, AB = 12 cm, AC = 20 cm and BD = 5 cm, determine CD.

It is given that $A D$ bisect $\angle A$. Also, $A B=12 \mathrm{~cm}, A C=20 \mathrm{~cm}$ and $B D=5 \mathrm{~cm}$.

We have to find $C D$.

Since $A D$ is the bisector of $\angle A$

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

$\frac{12 \mathrm{~cm}}{20 \mathrm{~cm}}=\frac{5 \mathrm{~cm}}{D C}$

$12 \mathrm{~cm} \times D C=20 \mathrm{~cm} \times 5 \mathrm{~cm}$

$D C=\frac{100}{12} \mathrm{~cm}$

$=8.33 \mathrm{~cm}$

Hence $C D=8.33 \mathrm{~cm}$