In an equilateral triangle, prove that the centroid and the circumcentre of the triangle coincide.

AD, BE and CF are medians of the triangle.

Let G be the centroid of triangle ABC.

Triangle ABC is an equilateral triangle,

∴ AB = BC = CA and ∠ABC = ∠BAC = ∠BCA = 60°

In △BFC and △BEC,

⇒ BC = BC (Common Side)

⇒ ∠FBC = ∠ECB = 60°.

⇒ BF = EC (As F is mid-point of AB and E is mid-point of AC and AB = AC.)

△BFC ≅ △BEC (By SAS axiom)

∴ BE = CF (By C.P.C.T.) …..(1)

Now, in △ABE and △ABD,

AB = AB (Common Side)

∠BAE = ∠ABD = 60°

BD = AE (As D is mid-point of BC and E is mid-point of AC and BC = AC.)

△ABE ≅ △ABD (By SAS axiom.)

∴ BE = AD (By C.P.C.T.) ……(2)

From equation 1 and 2, we get:

⇒ AD = BE = CF

⇒ $\dfrac{2}{3} AD = \dfrac{2}{3} BE = \dfrac{2}{3} CF$

We know that the centroid of the triangle divides the median in a 2 : 1 ratio.

∴ GA = GB = GC.

So, we can say that G is equidistant from the three vertices A. B and C.

G is circumcentre of ΔABC.

Hence, proved that the centroid and circumcentre are coincident.