#### Answer

4$\frac{8}{13}$ in

#### Work Step by Step

Given the triangle is right angles triangle and two sides are 5 in and 12 in
By using pythagoras theorem
$AC^{2}$ =$AB^{2}$ + $BC^{2}$
$AC^{2}$ = $5^{2}$ + $12^{2}$
$AC^{2}$ = 25 + 144
$AC^{2}$ = 169
AC = 13 in
Let us assume AD =x
AC = AD + DC
13 in = x + DC
DC = 13 - x
Now apply Pythagoras theorem on triangle BDC
$BC^{2}$ =$BD^{2}$ + $DC^{2}$
$12^{2}$ = $BD^{2}$ + $(13 - x)^{2}$
$BD^{2}$ = $12^{2}$ - $(13- x)^{2}$ - Eq 1
Similarly apply Pythagoras theorem on triangle ABD
$AB^{2}$ = $AD^{2}$ + $BD^{2}$
$5^{2}$= $BD^{2}$ + $x^{2}$
$BD^{2}$ = $5^{2}$- $x^{2}$ - eq2
$BD^{2}$ is common in both the equations So,
Eq1 = eq2
$12^{2}$ - $(13- x)^{2}$ = $5^{2}$- $x^{2}$
144 - 169- $x^{2}$ + 26x= 25 - $x^{2}$
-25 + 26x = 25
26x = 50
x = $\frac{50}{26}$
Now put the value of x in eq1
$BD^{2}$ = 25 - $\frac{50}{26}^{2}$
$BD^{2}$ = 25 - $\frac{2500}{676}$
$BD^{2}$ = $\frac{16900 - 2500}{676}$
BD = $\sqrt \frac{14400}{676}$
BD = $\frac{120}{26}$ in = 4$\frac{8}{13}$ in