Prove triangle ABC is equilateral.?

See graphic.

Construct equilateral triangle DEF, and circles centered at A, B, C, with radii = DB = EC = AF. Draw arbitrary AB passing through D, which represents a side of triangle ABC such that DB = AF. Draw OA, OB, where O is the center of DEF. Proving that ∠FAP = ∠DBO = 30 would lead to ΔABC being an equilateral.

I'll get back to this in a while, this proof is incomplete.

A weak proof would be as follows: Given arbitrary A on circle F, the line AD extended to circle D results in an unique point B, and the line BE extended to circle E results in an unique point C, which has to be the same unique point that results from the line AF extended to circle E. Thus, if given radii DB = EC = AF, there exists a solution set {A, B, C}, it has to be unique, and we know that since an equilateral triangle ΔABC would be a solution, it's the only one possible. But of course this is a very iffy "proof". Just leave this open.

Edit: Brian has just now sent me a proof that has already been worked out for this problem, see 2nd link. Yes, this is doing it in the hard way, you would think there'd be an easier way to prove this.

Edit 2: Analytically, given length DB = EC = AF for given size of △DEF, there's always at least 3 distinct solutions, not one, but only one meets the criteria that "D is between A & B, E is between B & C, and F is between C & A". It's notoriously difficult to distinguish "betweenishness" with points on a line, analytically speaking. See 3rd link for a typical pseudo solution that's hard to distinguish from the real thing analytically.

In △ABC

D is between A and B

E is between B and C

F is between C and A

DB = EC = AF

△DEF is equilateral.

DB = DE, EC = EF, AF = DF

All four triangles are equilateral.

Hence:

△ABC is equilateral.

by the transitive property of segment concruence you can connect multiple ideas, so i would use that.

If all sides in any triangle are equal in length, the triangle is automatically equiangular, thus making it equilateral. Welcome!