重心G = (A+B+C)/3,
　内心I = {(sinA)･A + (sinB)･B + (sinC)･C}/(sinA+sinB+sinC),
　外心O = {sin(2A)･A + sin(2B)･B + sin(2C)･C}/(sin(2A)+sin(2B)+sin(2C)),
　垂心H = {(tanA)･A + (tanB)･B + (tanC)･C}/(tanA+tanB+tanC)

∴　どれかが一致すれば正三角形ですね。

なお、G = (2O+H)/3　(オイラー線)