GIVEN: An equilateral triangle ABC, Medians AP, BQ, &CR. Their point of concurrency is O, which is the centroid of the triangle.

TO PROVE THAT: Centroid O is the circumcentre of the triangle ABC.

If we prove that centroid O is the circumcentre of the triangle, then it automatically becomes the centre of the circumcircle.

PROOF: Since AP is median, so P is mid point of BC.

ie, BP = PC.

AB = AC ( as triangle ABC is equilateral)

AP=AP ( common side)

Hence triangle ABP is congruent to ACP( by SSS congruence criterion)

=> angle APB = angle angleAPC ( corresponding parts of congruent triangles)

But their sum = 180°

So, each angle has to be 90°.

That shows that AP is perpendicular bisector of BC.

Similarly, pyove that BQ & CR are perpendicular bisectors of AC & AB respectively.

So now, The point of concurrency ‘O' of these perpendicular bisectors becomes circumcentre of the triangle. ( as circumcentre is the point of concurrency of 3 perpendicular bisectors of the sides of the triangle). And this centre is also the centre of circum circle.

This way centroid O coincides with circumcentre O…