Circle through equilateral triangle

Construct equilateral triangle ABC

Draw angle bisectors for A & B that cut BC & AC at F & E resp.    AF & BE intercept at D

     ∠EAD = 30∘ = ∠FBD      (construct)

     ∠EDA = ∠FDB            (opp. ang.s of intersecting str. lines)

     ∠DAB = 30∘ = ∠DBA      (construct)

   ∴ ∆DAB is isos.         (2 angles equal)

   ∴   AD = BD

   ∴ ∆EAD ≣ ∆FDB           (2 ang.s & inc. side =)

   ∴   ED = FD & EA = FB

But    AC = BC             (equilateral triangle)

   ∴   CE = CF             (equals minus equals are equal)

Join CD 

     ∆CED ≣ ∆CFD              (all 3 sides =)

   ∴ ∠ECD = ∠FCD

          = 30∘             (∠ECF = 60∘ as ∆ABC equilateral)

          = ∠EAD

   ∴ ∆ACD isos.

     ∴ CD = AD

       AD = DB             (proven)

So D is centre of circle of radius AD and passes through A, B & C.