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?

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.
