# Proving that DEF is also an equilateral triangle

The question gives that ABC is an equilateral triangle, AD = BE = CF and ∠DAB = ∠EBC.

I tried using only cases of triangle congruences to prove it, but I'm not sure if I did it correctly. Thus, take a look at my proof and tell me if everything is okay or not, please.

Here's my attempt:

Since AB = CB, BE = AD and ∠DAB = ∠EBC, then I have that ADB and BEC are congruent triangles by SAS.

• By the way, CF = BE (question gives) which is the same as CE + EF = BD + DE, but CE = BD, then EF = DE after replacing in formula.

(2 sides equals, lefting just one to prove that they are equal - in order to get the equilateral triangle)

Now, note that ∠A = ∠B = ∠C = ∠ABD + ∠EBC.

• However, since ∠DAB = ∠EBC, then I have ∠A = ∠B = ∠C = ∠ABD + ∠DAB.

• Cince triangle ADB and triangle BEC are congruent, it follows that ∠DBA = ∠ECB. So, ∠C = ∠DAB + ∠ECB.

• Looking at ∠C especially, we have ∠C = ∠DAB + ∠ECB. However, ∠C = ∠ACF + ∠ECB. So, ∠DAB + ∠ECB = ∠ACF + ∠ECB, then ∠DAB = ∠ACF.

• Now, since AC = AB, ∠DAB = ∠ACF and CF = AD, then ACF and BAD are congruent triangles by SAS congruence case.

• After that, I have the 3 triangles are congruent as well.

• So since AD = CF (question gives) and AF = CE (because AFC and , then for AD = AF + FD and CF = CE + EF, I have FD = EF.

Therefore, for EF = DE and FD = EF, it follows EF = DE = FD.

By the definition of equilateral triangle, I got that DEF is also an equilateral triangle!

I'm sorry for not only this huge proof, but also for not knowing how to use whether LaTeX or MathJax (I have bad skills) and I hope you understand that.

All help is appreciated!

• ACE = 60 – ECB = 60 – ADB = EBC. Similar arguments ( used in the first paragraph) can be applied to show ED = FD. – Mick Jun 11 '19 at 8:55
• Got it! Thanks @Mick – Daniel Sehn Colao Jun 11 '19 at 11:36

After finding $$\Delta DAB \cong \Delta CBE$$, this implies that $$\angle ABE = \angle BCF$$, you can simply use $$\angle ACF + \angle BCF = \angle ABE+ \angle CBE$$ and since $$\angle ABE = \angle BCF$$, we have $$\angle ACF = \angle CBE$$. From here, you can directly conclude $$\Delta ACF \cong \Delta CBE$$ by SAS since $$AC = BC$$, $$CF = BE$$ and $$\angle ACF = \angle CBE$$.
Then $$AF = CE = BD$$ and $$CF = BE = AD$$ imply $$EF = FE = ED$$, so the result follows.