an equilateral triangle with angle F

Let the length of each side of the triangle $\triangle {ABC}$ be $a$ , $|\overline {BD}|=|\overline {EF}|=x$ , $|\overline {AE}|=|\overline {CF}|=y$ , $\angle {BFD}=α$ .

Then $\dfrac {a-x}{\sin (60\degree-α)}=\dfrac {y}{\sin (60\degree+α)}$

$\dfrac {x}{\sin 60\degree}=\dfrac {y}{\sin (60\degree-α)}$

$\dfrac {a-y}{\sin α}=\dfrac {x}{\sin 60\degree}$

From these equations we get

$\sin α\sin (60\degree+α)+\sin (60\degree-α)\sin (60\degree+α)=\sin 60\degree\sin (60\degree+α)+\sin^2 (60\degree-α)$

$\implies 16\cos^4 α-8\cos^3 α-12\cos^2 α+8\cos α-1=0$

There are two acute angle solutions to this equation : $40\degree$ and $60\degree$ , of which the wanted solution is $\boxed {40\degree}$ .