Find ED and (I am so so sorry for my bad English in advance, it is my math homework problem, I tried to solve it but I've got no idea whether the solution's right or wrong:))) It is given that $∠B=120^{o}$, $AB=6 cm$, $BC=4cm$, $CE=BE$, $AC=CD$. Find the lenght of ED and angle $∠CED$. Look 
Solution:
$AC=\sqrt{6^2+4^2-2*6*4*\cos{120^o}}=2\sqrt{19}\;cm$
$\frac{6}{\sin(∠ACB)}=\frac{2\sqrt{19}}{\sin 60^o} \Rightarrow \sin(∠ACB)=\frac{3\sqrt3}{2\sqrt{19}}$
$∠ACB<90^o\Rightarrow  ∠DCB>90^o$
$\cos(∠DCB)=-\cos(∠ACB)=-\sqrt{1-(\frac{3\sqrt3}{2\sqrt{19}})^2}=\frac{-7}{2\sqrt{19}}$
$ED=\sqrt{(2\sqrt{19})^2+2^2+2*2\sqrt{19}*2*\frac{7}{2\sqrt{19}}}=6\sqrt{3} \;cm$
$\frac{6\sqrt{3}}{\frac{3\sqrt3}{2\sqrt{19}}}=\frac{2\sqrt{19}}{\sin(∠CED)} \Rightarrow \sin(∠CED)=\frac{1}{2} \Rightarrow ∠CED=30^o$
Is it correct guys? Thank you :)).

$$ED^2=4+CD^2-4CD\cos(180^{\circ}-\gamma)$$
$$CD^2=AC^2=36+16-48\cos(120^{\circ})$$
$$\frac{\sin(\gamma)}{\sin(120^{\circ})}=\frac{6}{AC}$$
$$AC^2=4+ED^2-4ED\cos(\delta)$$
