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?
 A: 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.
A: The proof is ugly due to the limitation of the Ecludean framework. I try to break it down into 3 diagrams.
See the first. As mentioned, $\triangle AEC$ is congruent to  $\triangle BDC$. [Proof is therefore skipped.]

Then $\alpha = \beta$ [result #1]
And BF = AE = BD [result #2]
See figure 2. Join EF. Through O, draw MN // AE cutting EF at N. HOK is similarly draw such that HOK // BF. [We have to assume MON and HOK are not the same straight line yet.]

By midpoint and intercept theorem, MO = HO = OK = ON = 0.5AE = 0.5BF.
Hence, ⊿MOH and ⊿NOK are congruent. This further implies AB // EF.
In addition, by midpoint theorem, EN = 0.5EF and FK = 0.5FE and therefore MON and HOK are actually the same straight line. And hence, ..., ABFE is a parallelogram.
See figure 3. α + 60 + 60 + γ = 180 [int. angles AE // BF]

Therefore, β + γ = 60 [from result #1]
The above plus [result #2] give you the required equilateral triangle.
A: There's an ambiguity to the problem, so I'll prove a converse and see how that might apply.

If $\square ABFE$ is a parallelogram, and $\triangle ABC$ and $\triangle BFD$ are equilateral triangles overlapping the parallelogram's interior, then $\triangle CDE$ is also equilateral.


A little angle-chasing provides us with the following:
$$\begin{align}
180^\circ &= \angle BAE + \angle ABE \\
&= \left(\;60^\circ + \theta \;\right) + \left(\;(60^\circ-\phi)+\phi+(60^\circ-\phi)\;\right) \\
&= 180^\circ + \left(\;\theta - \phi\;\right) \\
\implies\quad\theta &= \phi
\end{align}$$
Therefore, $\triangle CAE \cong \triangle EFD \cong \triangle CBD$ (by Side-Angle-Side), so that $\overline{EC}\cong\overline{ED}\cong\overline{CD}$, as claimed. $\square$

Now, the problem as originally stated fails to establish a unique point $F$. (See my comments to the question.) Undoubtedly, the intent is to have $\square ABFE$ be a parallelogram. If we take that as a premise, then $D$ is "the" point for which $\triangle CDE$ is an equilateral triangle; by the converse proven above, we know that the equilateral built on $\overline{BF}$ provides a point ---call it $D^\prime$--- such that $\triangle CD^\prime E$ is equilateral. There is obviously only one such point, so we conclude that $D = D^\prime$, and the problem is solved.
Except ... not really.
The point $D$ making $\triangle CDE$ equilateral is, in fact, not unique. There's a candidate for $D$ on each side of $\overline{CE}$. So, it's possible that the problem's $D$, and our $D^\prime$, are distinct points, so that the problem isn't solved. Nevertheless, one imagines that whatever semantics resolve the $F$-ambiguity in the problem also grant us the assurance that $D$ and $D^\prime$ lie on the same side of $\overline{CE}$ and must therefore coincide ... so perhaps the problem should be considered solved, after all.
It's hard to say without knowing exactly how one fixes the original problem.
The OP suggests requiring that $\triangle CDE$ and $\triangle ABC$ don't overlap. This seems to resolve the $F$-ambiguity. However, the converse result is true regardless of how the various components overlap with one another:

So, the proposed fix ignores the versatility of the underlying geometry. On the other hand, a fix that tries to accommodate this versatility could be overly-verbose. 
In any case, I find the converse to be the more-compelling aspect of the problem. I'll close by noting that the result holds even when the equilaterals are taken to overlap the parallelogram's exterior: 

This reminds me of external-vs-internal variants of Napoleon's Theorem.
A: Assuming that $BF\parallel AE$, here's a variation of @JackD'Aurizio's proof.
Rotate $\triangle CEA$ an angle $60^\circ$ counterclockwise about $C$, one gets $\triangle CDB$. Thus $AE=BD$ and $\angle(\vec{AE},\vec{BD})=60^\circ$. The conclusion follows from the fact that $\vec{AE}=\vec{BF}$.
