Who can do this Geometry Problem?

A concrete example :

In $\triangle {ABC}$ , $|\overline {AC}|=|\overline {BC}|=\sqrt 7, |\overline {BD}|=\dfrac{8}{\sqrt 7}, |\overline {BE}|=\dfrac{2}{\sqrt 7}, |\overline {AD}|=\dfrac{4}{\sqrt 7},|\overline {CD}|=\dfrac{3}{\sqrt 7}$ . Let $\angle {DEC}=α, \angle {CDE}=β$ . Then $\sin β=\sqrt 3\sin α, \sin (α+β)=2\sin α$ . This yields $\cos α=\dfrac{\sqrt 3}{2}\implies α=30\degree$ . So $\sin β=\sqrt 3\sin α=\dfrac{\sqrt 3}{2}\implies β=60\degree\implies β=2α$ . Hence $k=\boxed 2$