In $$\triangle ABC, \angle C = 90^\circ, AC = 5 cm$$ and $$BC = 12 cm$$. The bisector of $$\angle A$$ meets BC at D. What is the length of AD?

By the Pythagoras theorem,

$$(AB)^2 = (AC)^2 + (BC)^2$$

$$(AB)^2 = (5)^2 + (12)^2$$

$$(AB)^2 = (5)^2 + (12)^2$$

$$(AB)^2 = 25 + 144$$

AB = 13 cm

By angle bisector theorem,

$$\frac{AB}{BD} = \frac{AC}{CD}$$

Let CD be x cm.

$$\frac{13}{12 - x} = \frac{5}{x}$$

13x = 60 - 5x

x = 60/18 = 10/3

In $$\triangle ACD$$,

$$(AD)^2 = (AC)^2 + (CD)^2$$

$$(AD)^2 = (5)^2 + (\frac{10}{3})^2$$

$$(AD)^2 = 25 + \frac{100}{9}$$

$$(AD)^2 = \frac{325}{9}$$

$$AD = \frac{5\sqrt13}{3}$$