Question

Prove that in an equilateral triangle, the circumcentre and incentre are the same. What is the ratio of the circumradius and inradius?

Answer 1

Case I.

Consider an equilateral triangle ABC.

Let O′ be its incentre.

We know that incentre of a circle is obtained by the intersection of the angles bisectors of a triangle.

BO′ is the angle bisector of ∠ABC.

∴ ∠ABO′ = ∠O′BC =

Similarly, we have ∠O′CA = ∠O′AC = ∠O′AB = ∠O′BA = 30°

Case II.

Now, again consider the same triangle with angles marked on it.

Let O be the circumcentre of the given triangle.

As O is the circumcentre of ΔABC, OA = OB = OC

In ΔOBC, OB = OC

⇒ ∠2 = ∠1 …(1)

Also, ∠1 + ∠6 = ∠2 + ∠3 (As each angle of an equilateral triangle is equal)

⇒ ∠6 = ∠3 …(2)

Similarly, in ΔOAB:

∠5 = ∠6 …(3)

In ΔOAC:

∠3 = ∠4 …(4)

From equations (2) and (4), we have:

∠4 = ∠6 …(5)

From equations (3) and (5), we have:

∠4 = ∠5

As ∠4 + ∠5 = 60°

⇒ 2∠4 = 60°

⇒ ∠4 = 30°

Similarly, we can show that ∠1 = ∠2 = ∠3 = ∠4 = ∠5 = ∠6 = 30°, i.e., ∠ABO = ∠OBC = ∠OCA = ∠OAC = ∠OAB = ∠OBA = 30°.

Therefore, BO, AO and CO is the angle bisectors of the angles of the given triangle ABC.

However, BO′, AO′ and CO′ is the angle bisectors of the angles of the given triangle ABC according to our assumption in case I.

Thus, O and O′ coincide.

Hence, the circumcentre and incentre of an equilateral triangle are same.

We know that centroid of an equilateral triangle coincides with circumcentre and orthocentre.

Also, centroid divides the median in the ratio 2:1.

Here, AD is the median.

⇒ AO = and OD =

⇒ AD = 3 OD

⇒ AO = 2 OD

⇒ AO : OD = 2 : 1

Thus, the ratio of the circumradius to inradius is 2:1.