Square and triangles in a simple geometry problem Given $ABCD$ is a square with side "$a$"  and  both $\triangle ABE$ and $\triangle BCF$ are equilateral triangles, how can I find the area of $\triangle BEF$ ?       

I know that $BE=BF=a$ ... but I would need $EF$ for the perimeter formula for the area .      

The situation is represented here :     



Thanks a lot!
Hints:

$$\angle EBF=90^\circ\implies \;\Delta BEF\;\;\text{is an isosceles straight-angle triangle}\implies$$

$$EF=\frac2{\sqrt2}a=\sqrt2\,a$$
Isn't it easier to divide the area of the square $BEDF$ in two?