Question

In an equilateral triangle, prove that the centroid and the centre of the circumcircle (circumcentre) coincide. [4 MARKS]

Answer 1

Concept: 2 Marks
Application: 2 Marks

Given: In an equilateral triangle ABC, D, E, F are the mid-points of sides BC, CA and AB respectively.

Proof: In $\Delta ABC$, let G be the centroid.

G is the point of intersection of AD, BE and CF.

In $\Delta BEC$ and $\Delta BFC$

$BC=BC$ [common]

$\angle B=\angle C={60}^{\circ }$ [equilateral triangle]

$BF=CE$

$[BF=\frac{AB}{2}=\frac{AC}{2}=CE]$

$\Delta BEC\cong \Delta BFC$ [S.A.S Congruency]

$BE=CF$ ..........(1) [C.P.C.T]

Similarly, in congruent triangles $\Delta CAF$ and $\Delta CAD$, we get

$CF=AD$ .........(2) [C.P.C.T]

From (1) and (2), we get

$BE=CF=AD$

$\frac{2}{3}BE=\frac{2}{3}CF=\frac{2}{3}AD$

$GB=GC=GA$

G is equidistant from the vertices

G is circumcenter of $\Delta ABC$

The centroid and circumcenter are coincident.