find the length of fb

If the angle is $\theta$ , then $ab = 3 \tan 2 \theta$ , and $ac = 3 \tan 2 \theta \tan \theta$

hence $cd = 3 \tan 2 \theta \tan \theta - 3$

$fb = bd - fd = bd - cd = \dfrac{3}{ \cos 2 \theta} - 3 \tan 2 \theta \tan \theta + 3$

$= \dfrac{3}{ \cos 2 \theta} ( 1 - \sin 2 \theta \tan \theta + \cos 2 \theta )$

$= \dfrac{3}{ cos 2 \theta } ( 1 - \dfrac{2 \sin^2 \theta cos \theta } { cos \theta } + \cos 2 \theta )$

$=\dfrac{3}{ cos 2 \theta } ( 1 - 2 \sin^2 \theta + \cos 2 \theta )$

$=\dfrac{3}{ cos 2 \theta } ( \cos 2 \theta + \cos 2 \theta ) = \boxed{6}$