<s>[INST] 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.

 [/INST] 
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 $\gamma=\frac{\pi}{2}-\frac{\alpha}{2} \implies \tan BAD=-\tan(\pi-\alpha)=\tan \alpha=\frac{\sqrt{55}}{3}$
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