Let us prove that in an equilateral triangle, circumcentre, incentre, centriod, orthocenter will coincide.

We know,

Circumcenter of a triangle: The intersection of perpendicular bisectors of the triangles.

Incenter: The intersection of angle bisectors of the triangle.

Centroid: The intersection of the medians of the triangle.

Orthocentr: The intersection of the altitudes of the triangle

In order to prove that, In an equilateral triangle, circumcenter, incentre, centroid and orthocenter coincide, it is sufficient to prove that for any side median, altitude, perpendicular bisector and angle bisector of angle opposite to that side is common.

Let us consider an equilateral ΔABC, such that AD is a median to side BC.

To Prove:

(i) AD ⊥ BC [AD is Altitude and perpendicular bisector of BC]

(ii) ∠BAD= ∠CAD [AD is angle bisector of ∠A]

Proof:

In ΔABD and ΔACD

AB = AC [Sides of equilateral triangle]

AD = AD [Common side]

BD = CD [As, AD is median to BC]

⇒ ΔABD ≅ ΔACD [By SSS property of congruent triangles]

Now, As corresponding parts of congruent triangles are equal [CPCT], we have

∠ADB = ∠ADC

Also,

∠ADB + ∠ADC = 180° [Linear pair]

⇒ ∠ADB + ∠ADB = 180°

⇒ ∠ADB = ∠ADC = 90°

⇒ AD ⊥ BC

Since, AD ⊥ BC, and BD = BC

∴ AD is perpendicular bisector of BC and as well as altitude from A to BC.

Now,

∠BAD = ∠CAD [By CPCT]

AD is angle bisector of ∠A.

∴ AD is median, perpendicular bisector, altitude and angle- bisector.

We can prove this result for every median, and since lines are same, their intersection will also be same.

Hence Proved!