A $\triangle ADE$ has $\angle AED = 120^\circ$, and $B$ and $C$ are points on $AD$ such that $\triangle BCE$ is equilateral. If $AB = 4$ cm, $CD = 16$ cm and $BC = x$ cm, find $x$ .
What I Tried: Here is a picture :-
No idea for this problem. I could proceed by angle-chasing, considering $\angle BEA = \theta$ , but I don't think that will help me in finding $BC$ without Trigonometry. I think I need to use Law of Cosines on $\angle AEC$ to solve this, but I am new to it and I prefer simple geometry techniques. There are nothing I can use such as similarity, Pythagoras theorem or areas, etc.
Can anyone help me? Thank You.
Best Answer
Notice $\angle ECD = 120 = \angle ABE$. In $\triangle AED$, one angle is $120$ and it shares one angle each with smaller unshaded triangles.
We conclude
$$ \triangle AED \sim \triangle ABE \sim \triangle ECD$$
by AA similarity.
Thus $\triangle ABE \sim \triangle ECD$ and $$\dfrac{AB}{BE}=\dfrac{EC}{CD}$$