In an equilateral triangle prove that the centroid and the centre of the circumcircle (circumcentre) coincide.

Explanation

Given: An equilateral triangle ABC in which D,E and F arc the mid-points of sides BC, CA and AB respectively.
To Prove : The centroid and circumentre are coincident.
Construction : Draw medain AD.
Proof : In `Delta ADB` and `Delta ADC`.
`because" "{(AB,=,,AC,("sides of an equilateral triangle")),(AD,=,,AD,("common")),(BD,=,,DC,(because"D is the mid-points")):}`
`thereforeDelta ADBcongDeltaADC` (SSS congruency)
`thereforeangle1=angle2`(c.p.c.t)
But `angle1+angle2=180^@` (L.P.A.)
`thereforeangle1=angle2=90^@`
and `BD=DC` (given)
`therefore` AD is the perpendicular bisects of BC.
It means median and perpendicular bisector are same in an eqilateral triangle.
`rArr` All the medians =All the perpendicular bisectors
`rArr` Intersection point of medians =Intersection point of perpendicular bisectors.
`rArr` Centroid = Circumcentre.
i.e., centroid and circumcentre coincide in an equilateral triangle.