PROVE TRIANGLE ABC IS EQUILATERAL.?

I have a solution, but you may find it unsatisfactory.

Begin with an equilateral triangle DEF. Through point F construct a line ℓ at an angle θ with FD. On the side of ℓ closer to D, construct the point A at the given distance AF (=DB=EC) from F.

Now, construct the line AD, and construct the point B at the given distance DB (=AF=EC) from D on the opposite side as A.

Finally, construct the line BE, and let C be its intersection with line AF (i.e. line ℓ). Now, depending on θ, we might or might not have EC equal to the prescribed length (i.e. EC=DB=FA). In fact, by continuity there is only (or at most) one unique θ so that EC=DB=FA.

But it's fairly easy to see that if any solution exists for a given DEF and AF, then an equilateral solution exists; therefore the only solution (if any) is equilateral. In other words, ABC must be an equilateral triangle.

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