Construction of an equilateral triangle from two equilateral triangles with a shared vertex

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.

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.