Question

In an equilateral triangle, prove that the centroid and center of the circum-circle (circumcentre) coincide.

Answer 1

Given: An equilateral triangle ABC in which D, E, and F are the midpoints of sides BC, CA and AB respectively.

To prove: The centroid and circumference are coincident.

Construction: Draw medians AD, BE and CF.

Proof:
Let G be the centroid of ΔABC i.e., the point of intersection of AD, BE, and CF. In triangles BEC and BFC, we have
∠ B = ∠ C = 60°
BC = BC
and BF = CE    ...[ ∵ AB = AC ⇒ ${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}}$ AC ⇒ BF = CE ]
∴ ΔBEC = ΔBFC
⇒ BE = CF           ...(i)
Similarly,
Δ CAF and Δ CAD
⇒ CF = AD           ...(ii)

From (i) and (ii),
AD = BE = CF
⇒ ${\displaystyle \frac{2}{3}\text{AD}=\frac{2}{3}BE=\frac{2}{3}\text{CF}}$

CG = ${\displaystyle \frac{2}{3}\text{CF}}$

GA = ${\displaystyle \frac{2}{3}\text{AD}}$

GB = ${\displaystyle \frac{2}{3}\text{BE}}$

GA = GB = GC

⇒ G is the equidistant from the vertices
⇒ G is the circumcentre of ΔABC.
Hence, the centroid and circumcentre are coincident.