In $\triangle ABC, \angle C = 90^\circ$ and D is a point on CB such that AD is the bisector of $\angle A$. If AC = 5 cm and BC = 12 cm, then what is the length of AD?

$\frac{5\sqrt{13}}{3}$ cm

BC = 12, AC = 5

In \(\Delta \)ABC

\( { (AB)}^{2 } \) = \( { (AC)}^{2 } \) + \( { (BC)}^{2 } \)

\( { (AB)}^{2 } \) = \( { (5)}^{2 } \) + \( { (12)}^{2 } \) = 25 + 144 = 169

AB = \(\sqrt {169 }\) = 13

As AD is the bisector of \(\angle\)A

= \(\frac{AC}{AB}\) = \(\frac{CD}{DB}\)

= \(\frac{5}{13}\) = \(\frac{CD}{DB}\)

Suppose CD : DB = 5x : 13x

BC = CD + DB

12 = 5x + 13x

18x = 12

x = \(\frac{12}{18}\) = \(\frac{2}{3}\)

CD = 5 x \(\frac{2}{3}\) = \(\frac{10}{3}\)

In \(\Delta \)ACD

\( { (AD)}^{2 } \) = \( { (CD)}^{2 } \) + \( { (AC)}^{2 } \)

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

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

\( { (AD)}^{2 } \) = \(\frac{100\;+\;225}{9}\) = \(\frac{325}{9}\)

AD = \(\frac{5\sqrt {13 }}{9}\).

Therefore, AD is \(\frac{5\sqrt {13 }}{9}\) cm.