if $AB=AC=EF,AE=BE,\angle BAC=120^{\circ}$ then find $\angle ABE$ If $$AB=AC=EF, AE=BE, \angle BAC=120^{\circ}, \angle ADB=\angle BEF=90^{\circ}$$
find $\angle ABE$

Following is one method:
Let $\angle ABE=a, \angle AFE=\angle DBE=30^{\circ}-a, \angle EAF=60^{\circ}-a$. Use sine thereom, we have
$$\dfrac{\sin{a}}{\sin{(180^{\circ}-2a)}}=\dfrac{AE}{AB}=\dfrac{AE}{EF}=\dfrac{\sin{(30^{\circ}-a)}}{\sin{(60^{\circ}-a)}}$$
$$\Longrightarrow 2\cos{a}\sin{(30^{\circ}-a)}=\sin{(60^{\circ}-a)}$$
$$\Longrightarrow  \dfrac{1}{2}+\sin{(30^{\circ}-2a)}=\sin{(60^{\circ}-a)}$$
$$\Longrightarrow 2\cos^2{(a+30^{\circ})}-\cos{(a+30^{\circ})}-\dfrac{1}{2}=0$$
$$\Longrightarrow \cos{(a+30^{\circ})}=\dfrac{\sqrt{5}+1}{4}=\cos{36^{\circ}}$$
so we have
$$a=6^{\circ}$$
Question: Can  find geometry methods or other simple methods?


Let $\angle EBA = \alpha$. 
Let $A$ and $X$ be symmetric with respect to $BE$. Then $BX=AB$. Also note that $BE=AE=EX$, so $A,B,X$ lie on the circle with center $E$ and radius $AE$.
Let $Y$ be on $AF$ and $AY=AB$. Since $\angle BAY = 60^\circ$, $\triangle BAY$ is equilateral. Thus $BY=AB$. Therefore $A,X,Y$ lie on the circle with center $B$ and radius $AB$.
Since $AE=BE$ and $AY=BY$, triangles $AYE$, $BYE$ are congruent (sss). In particular $\angle AYE = \angle EYB = \frac 12 \angle AYB = 30^\circ$. 
We have 
$$\angle AXY = \frac 12 (360^\circ - \angle YBA) = 150^\circ = 180^\circ - \angle AYE = \angle EYF.$$
Moreover
$$\angle XYA = \frac 12 \angle XBA = \alpha$$
and
$$\begin{align*}\angle FEY & = \angle BEY - \angle BEF = \\
& = (180^\circ - \angle YBE - \angle EYB) - 90^\circ = \\ 
& = 180^\circ - (60^\circ - \alpha) - 30^\circ - 90^\circ = \\ 
& = \alpha,
\end{align*}$$
therefore
$$\angle XYA = \angle FEY.$$
It follows that triangles $AXY$ and $FYE$ are similar. Their similarity ratio is $\dfrac{AY}{EF} = 1$ so in fact they are congruent. Thus $EY=XY$. 
Using the fact that $EXY$ is isosceles we get
\begin{align*}180^\circ & = \angle XYE + 2 \angle YEX = \\
& = 30^\circ + \alpha + 2(\angle YEA - \angle XEA) = \\
& = 30^\circ + \alpha + 2\angle BEY - 2\angle XEA = \\
& = 30^\circ + \alpha + 180^\circ + 2 \alpha - 8 \alpha = \\
& = 210^\circ - 5 \alpha,
\end{align*}
therefore $$\alpha=6^\circ.$$
