#### Answer

12 in

#### Work Step by Step

Lets assume AD = x,
So AC = AD+ DC
14 = x +DC
DC = 14 - x
Now apply Pythagoras theorem on triangle ABD
$AB^{2}$ = $AD^{2}$ + $BD^{2}$
$13^{2}$ = $x^{2}$ + $BD^{2}$
$BD^{2}$ = $13^{2}$ - $x^{2}$ - Eq 1
Similarly apply Pythagoras theorem on triangle BCD
$BC^{2}$ = $CD^{2}$ + $BD^{2}$
$15^{2}$ = $(14 - x)^{2}$ + $BD^{2}$
$BD^{2}$ = $15^{2}$ - $(14 - x)^{2}$ - eq2
$BD^{2}$ is common in both the equations So,
Eq1 = eq2
$13^{2}$ - $x^{2}$ = $15^{2}$ - $(14 - x)^{2}$
169 - $x^{2}$ = 225 -196 - $x^{2}$ + 28x
169 = 29 + 28x
28x = 169 - 29
28x = 140
x = 5
Now put the value of x in eq1
$BD^{2}$ = $13^{2}$ - $5^{2}$
$BD^{2}$ = 169 - 25
$BD^{2}$ = 144
BD = 12 in