#### Answer

8 cm

#### Work Step by Step

Lets assume AD = x,
So AC = AD+ DC
21 = x +DC
DC = 21 - x
Now apply Pythagoras theorem on triangle ABD
$AB^{2}$ =$AD^{2}$ + $BD^{2}$
$10^{2}$ = $^{2}$ + $BD^{2}$
$BD^{2}$ = $10^{2}$- $x^{2}$ - Eq 1
Similarly apply Pythagoras theorem on triangle BCD
$BC^{2}$ = $CD^{2}$ + $BD^{2}$
$17^{2}$= $(21-x)^{2}$ + $BD^{2}$
$BD^{2}$ = $17^{2}$- $(21-x)^{2}$ - eq2
$BD^{2}$ is common in both the equations So,
Eq1 = eq2
$10^{2}$ - $x^{2}$ = $17^{2}$ - $(21-x)^{2}$
100 - $x^{2}$ = 289 -441 - $x^{2}$+ 42x
100 = -152 + 42x
42x = 252
x = 6
Now put the value of x in eq1
$BD^{2}$ = 100 - 36
$BD^{2}$ = 64
BD = 8 cm