Question

In the figure below, G is the centroid of ΔABC.

Prove that the triangles AGB, AGC, BGC have the same area.

Answer 1

Given: G is the centroid of ΔABC.

Construction: Extend AG such that it intersects BC at point D.

In ΔABD and ΔACD:

BD = DC (As AD is the median of the triangle)

Also, the height of both the triangles is equal.

∴ Area of ΔABD = × Height × BD

= × Height × DC

= Area of ΔACD

Area of ΔABD = Area of ΔACD …(1)

Now, in ΔGBD and ΔGCD:

BD = DC (As AD is the median of the triangle)

Also, the height of both the triangles is equal.

∴ Area of ΔGBD = × Height × BD

= × Height × DC

= Area of ΔGCD

Area of ΔGBD = Area of ΔGCD …(2)

Subtracting equation (2) from equation (1):

Area of ΔABD − Area of ΔGBD = Area of ΔACD − Area of ΔGCD

⇒ Area of ΔAGB = Area of ΔAGC

Similarly, Area of ΔAGB = Area of ΔGBC

⇒ Area of ΔAGB = Area of ΔGBC = Area of ΔAGC