Rhombus rotates in a circle. Two vertexes of the rhombus is on the circle, rotate the rhombus $ABCD$ clockwise around the point A to the rhombus $AB'C'D'$, where the point $B'$ falls on the circle, link $B'D,C'C$.
If $B'D:CC'=4:3$, then value of $tan\angle BAD$ will be?
It seems $B',B,D'$ or $D,D',C'$ are in a straight line on the image, but I can't proof it.By the way, there are also many things that seem right but hard to proof.It confuses me a lot.


If the angle of rotation is $\phi$ (clockwise) then
$ CC' = 2 \overline{AC} \sin(\dfrac{\phi}{2}) $
and
$ BD' = 2 \overline{AD} \sin(\angle BAC + \dfrac{\phi}{2} ) $
Now, $ \dfrac{\overline{AD}}{\overline{AC}} = \dfrac{1}{2} \sec(\angle BAC) $
Also, from the figure, we have $\phi =  \pi - 4 \angle BAC $
Let $ x = \angle BAC $ , then
$\dfrac{4}{3} = \dfrac{1}{2} \sec(x) \dfrac{ \sin( \dfrac{\pi}{2} -  x ) }{\sin( \dfrac{\pi}{2} - 2 x)}  $
Hence,
$8 \cos(2x) = 3 $
Therefore
$x = \angle BAC = \dfrac{1}{2} \cos^{-1}\left(\dfrac{3}{8}\right) $
Thus
$ \angle BAD = 2 x = \cos^{-1} \left( \dfrac{3}{8} \right ) $
Thus $ \cos(\angle BAD) = \dfrac{3}{8} $
From which
$\tan(\angle BAD) = \sqrt{\dfrac{64}{9} - 1} = \dfrac{\sqrt{55}}{3}$
The figure below shows the original rhombus (blue) and its rotated image (red).

