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In an equilateral triangle, prove that the centroid and center of the circum-circle (circumcentre) coincide.

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#### Solution

**Given: **An equilateral triangle ABC in which D, E, and F are the midpoints of sides BC, CA and AB respectively.

**To prove:** The centroid and circumference are coincident.

**Construction: **Draw medians AD, BE and CF.

**Proof:**

Let G be the centroid of ΔABC i.e., the point of intersection of AD, BE, and CF. In triangles BEC and BFC, we have

∠ B = ∠ C = 60°

BC = BC

and BF = CE ...[ ∵ AB = AC ⇒ `1/2"AB" = 1/2` AC ⇒ BF = CE ]

∴ ΔBEC = ΔBFC

⇒ BE = CF ...(i)

Similarly,

Δ CAF and Δ CAD

⇒ CF = AD ...(ii)

From (i) and (ii),

AD = BE = CF

⇒ `2/3"AD" = 2/3 BE = 2/3"CF"`

CG = `2/3"CF"`

GA = `2/3"AD"`

GB = `2/3 "BE"`

GA = GB = GC

⇒ G is the equidistant from the vertices

⇒ G is the circumcentre of ΔABC.

Hence, the centroid and circumcentre are coincident.

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