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.