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?

$\frac{5\sqrt{13}}{3}$ 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 = \(\frac{5\sqrt {13 }}{3}\).

Therefore, AD is \(\frac{5\sqrt {13 }}{3}\) cm.