Trigonometry/Geometry triangle help In right triangle $ABC$, ($\angle BAC = 90$), $D$ is found on side $AC$, $BD$ is the angle bisector of $\angle ABC$ and $BC= \sqrt3 BD$.

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*Find the value of $\angle CBD$.


*With regards to the point $E$ (found on $AC$), establish whether the areas of the circles which circumscribe the triangles $BDE$ and $BCE$ are equivalent.


*(Not connected to number 2): Given that $CE=2DE$, find the value of the angle $\angle BEC$.

Note: I tried to call $\angle ABC =\alpha$ and thereby bisect it. I found different trigonometric relationships, but I could not work out how to solve for either the values of the sides nor the angles.

(1) By sine law,
$${\sin({\pi\over 2}-x)\over \sin({\pi\over 2}-2x)}=\sqrt3$$
$$\sin({\pi\over 2}-x)- \sqrt{3}\sin({\pi\over 2}-2x)=0$$
$$\cos(x)-\sqrt{3}\cos(2x)=0$$
$$\cos(x)-\sqrt{3}(2\cos^2(x)-1)=0$$
$$6\cos^2(x)-\sqrt{3}\cos(x)-3=0$$
$$\cos(x)={\sqrt{3}\pm\sqrt{75}\over 12}={\sqrt{3}\pm5\sqrt{3}\over 12}={1\over2}\sqrt{3},-{1\over3}\sqrt{3}$$
Since $0<x<90$ degrees, we take the positive result and $x$ is therefore $30$ degrees.
(2) $BC>BE>BD$ as given.
Angle $C$ correspond to a larger arc in the circumcircle of $\triangle BCE$ than in $\triangle BCD$. This means the circumcircle of $\triangle BCE$ is larger than the circumcircle of $\triangle BCD$.
Angle $BDC$ correspond to a larger arc in the circumcircle of $\triangle BCD$ than in $\triangle BDE$. This means the circumcircle of $\triangle BCD$ is larger than the circumcircle of $\triangle BDE$.
Combining both results, the circumcircle of $\triangle BCE$ is larger than the circumcircle of $\triangle BDE$.
(3) From (1) we already knows $\triangle ABC$ is a $30,60,90$ triangle. This means $CD=2AD$ therefore $CD:DE:AD=4:2:3$. Since $AC:AB=\sqrt{3}$ we know $AE:AB={5\over 9}\sqrt{3}$ and $AB:AE={3\over 5}\sqrt{3}$. Therefore $\angle AEB=\arctan({3\over 5}\sqrt{3})$ and $\angle BEC=\pi - \arctan({3\over 5}\sqrt{3})$
