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.

 A: I find a solution without trigonometry. 

Link $CB'$ and $DC'$,$$\because DC=AB' \therefore \angle DB'C=\angle ADB'\\$$
Thus, $CB' \parallel AD$ 
Since, $CB \parallel AD$
$C,B,B'$ are in a straight line.
Similarly, $D,D',C'$ are also in a straight line.
Then, draw a line segment $B'H$ perpendicular to DC' which foot is point $H$.
Let the intersection of $CB'$ and $DC'$ be G.
Obviously, $$\triangle DGB'\sim \triangle CGC', $$and
$$\frac{B'G}{CG}=\frac{DB'}{CC'}=\frac{4}{3}$$
Hence $\angle AB'C= \angle DC'B' = \angle B'GC'$
Thus $B'G=B'C$
So $HC=\frac{1}{2}GC'$
Finally,  $tan \angle BAD = tan \angle B'C'D= \frac{B'H}{HC'}=\frac{\sqrt{55}}{3}$

A: The congruent sides of the rhombus and its rotation ensure that arcs $\stackrel{\frown}{CD}$, $\stackrel{\frown}{DA}$, $\stackrel{\frown}{AB'}$, $\stackrel{\frown}{B'C'}$ are all congruent, so they subtend congruent inscribed angles (such as $\angle CAD$), which are half the size of target angle $\angle BAD$.

Since chords $\overline{B'D}$, $\overline{AC}$, and $\overline{AC'}$ each span two of these arcs, they are congruent. Moreover, since inscribed angle $\angle ACC'$ subtends two of these arcs, it is congruent to $\angle BAD$.
Thus, with $M$ the midpoint of $\overline{CC'}$ (and necessarily the foot of the altitude from $A$), we can write
$$\tan\angle BAD = \tan\angle ACC'=\frac{|AM|}{|CM|}=\frac{\sqrt{|AC|^2-|CM|^2}}{|CM|}=\frac{\sqrt{|B'D|^2-\frac14|CC'|^2}}{\frac12|CC'|}$$
If $|B'D|:|CC'|=p:q$, this becomes
$$\tan\angle BAD=\frac{\sqrt{4p^2-q^2}}{q}\quad\stackrel{(p,q)=(4,3)}{=}\quad\frac{\sqrt{55}}{3}$$
which agrees with other solutions given. $\square$
A: 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).

A: 
Let $\angle AOD=\angle DOC=\alpha, \angle EOC=\beta, \angle ODC=\gamma$
We have that $\sin \alpha: \sin \beta=4:3$ or $3\sin \alpha=4\sin \beta$, also $\sin(2\alpha+\beta)=0$Thus,
$$\sin 2\alpha\cos \beta+\cos2 \alpha\sin\beta=0$$
$$2\sin \alpha\cos \alpha \cos \beta+\frac{3}{4}\cos2 \alpha\sin\alpha=0$$
$$2\cos \alpha \cos \beta+\frac{3}{4}\cos2 \alpha=0$$
Using that: $9\sin^2\alpha=16\sin^2\beta \implies \cos \beta=\frac{\sqrt{7+9\cos^2\alpha}}{4}$
$$\cos \alpha \frac{\sqrt{7+9\cos^2\alpha}}{2}+\frac{3}{4}(2\cos^2 \alpha-1)=0$$Thus, we have the equation to find $x=\cos \alpha$: $$x\sqrt{7+9x^2}+3x^2=\frac{3}{2}$$
$$x^2(7+9x^2)=\frac{9}{4}-9x^2+9x^4$$
$$16x^2=\frac{9}{4} \implies x=\frac{3}{8}$$ Next, notice that we need to find $\tan BAD=\tan (\pi-2\gamma)=-\tan 2\gamma$ and $\gamma=\frac{\pi}{2}-\frac{\alpha}{2} \implies \tan BAD=-\tan(\pi-\alpha)=\tan \alpha=\frac{\sqrt{55}}{3}$
