Trigonometry/Geometry triangle help

Question

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

1. Find the value of $\angle CBD$.

2. 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.

3. (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.

Answer 1

(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 $\angle BEC=\pi - \arctan({3\over 5}\sqrt{3})$

Answer 2

1.$AB=BD\cos(\alpha)\hspace 1cmAC=BC\sin(2\alpha)=\sqrt3BD\sin(2\alpha)\hspace 1cm BC=\sqrt3 BD$

this implies $2\sqrt3\cos(\alpha)^2-\cos(\alpha)-\sqrt3=0\Rightarrow \cos(\alpha)=\dfrac{\sqrt3}{2}$. Then $\angle{CBD}=30^{\circ}$

2.The two respective radius should be equal. From the two areas $$A=\frac{BD\cdot DE\cdot BE}{4R}\\A'=\frac{BE\cdot EC\cdot BC}{4R'}$$ we have the condition $$R=R'\iff(BD\cdot DE)A'=(BC\cdot EC)A$$ This can be set straightforward but it is some tedious.

3. Note that $\triangle{BAD}$ is the half of an equilateral triangle so $\angle{BDC}=120^{\circ}$ and $\angle{BCD}=30^{\circ}$, etc.

You can take my answer as a simple comment if you consider it as incomplete.