Construction of an equilateral triangle from two equilateral triangles with a shared vertex Problem
Given that $\triangle ABC$ and $\triangle CDE$ are both equilateral triangles. Connect $AE$, $BE$ to get segments, take the midpoint of $BE$ as $O$, connect $AO$ and extend $AO$ to $F$ where $|BF|=|AE|$. How to prove that $\triangle BDF$ is a equilateral triangle.

Attempt


*

*I've noticed that $\triangle BCD \cong \triangle ACE$ so that $|AE|=|BD|$, but I was stuck when proving $\triangle AOE \cong \triangle FOB$.

*Even though I assume that $ABFE$ is a parallelogram, I can neither prove one of three angles of $\triangle BDF$ is 60 degree nor prove $|DB|=|DF|$ through proving $\triangle DEF \cong \triangle DCB$ (which I think is right.)


Could anybody give me a hand?

Embed the construction in the complex plane.
Let $\omega=\exp\left(\frac{\pi i}{3}\right),B=0,C=1,E=1+v$.
Then $A=\omega$ and $D=1+\omega v$, hence $F=B+E-A$ implies:
$$ F = 1-\omega+v,$$
hence:
$$ \omega F = \omega -\omega^2 + \omega v = 1+\omega v = D, $$
so $BFD$ is equilateral.
