Question

If the medians of a ∆ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = `1/3` ar (ABC)

Answer 1

Given: In ΔABC, AD, BE and CF are medians and intersect at G.

To prove: ar (ΔAGB) = ar (ΔAGC) = ar (ΔBGC) = `1/3` ar (ΔABC)

Proof: We know that, a median of a triangle divides it into two triangles of equal area.

In ΔABC, AD is a median

∴ ar (ΔABD) = ar (ΔACD)  ...(i)

In ΔBGC, GD is a median

∴ ar (ΔGBD) = ar (ΔGCD)  ...(ii)

On subtracting equation (ii) from equation (i), we get

ar (ΔABD) – ar (ΔGBD) = ar (ΔACD) – ar (ΔGCD)

⇒ ar (ΔAGB) = ar (ΔAGC)  ...(iii)

Similarly, ar (ΔAGB) = ar (ΔBGC) ...(iv)

From equations (iii) and (iv),

ar (ΔAGB) = ar (ΔBGC) = ar (ΔAGC)  ...(v)

Now, ar (ΔABC) = ar (ΔAGB) + ar (ΔBGC) + ar (ΔAGC)

⇒ ar (ΔABC) = ar (ΔAGB) + ar (ΔAGB) + ar (ΔAGB)  ...[From equation (v)]

⇒ ar (ΔABC) = 3 ar (ΔAGB)

⇒ ar (ΔAGB) = `1/3` ar (ΔABC)  ...(vi)

From eqattions (v) and (vi),

ar (ΔBGC) = `1/3` ar (ΔABC)

And ar (ΔAGC) = `1/3` ar (ΔABC)

Hence proved.