Question

Question 8
If the medians of a $\Delta ABC$ intersect at G, then show that $ar\left(\Delta AGB\right)=ar\left(\Delta BGC\right)=ar\left(\Delta AGC\right)=\frac{1}{3}ar\left(\Delta ABC\right)$.

Answer 1

We know that, a median of a triangle divides it into two triangles of equal area.
In$\Delta ABC$, AD is a median.
$\therefore ar\left(\Delta ABD\right)=ar\left(\Delta ACD\right)$ ....(i)
In $\Delta BGC$, GD is a median.
$\therefore ar\left(\Delta GBD\right)=ar\left(\Delta GCD\right)$ ....(ii)
On subtracting Eq. (ii) from Eq. (i), we get,
$ar\left(\Delta ABD\right)-ar\left(\Delta GBD\right)=ar\left(\Delta ACD\right)-a\left(\Delta GCD\right)$
$\Rightarrow ar\left(\Delta AGB\right)=ar\left(\Delta AGC\right)$ ...(iii)
Similarly,
$ar\left(\Delta AGB\right)=ar\left(\Delta BGC\right)$ ...(iv)
From Eqs. (iii) and (iv),
$ar\left(\Delta AGB\right)=ar\left(\Delta BGC\right)=ar\left(\Delta AGC\right)$ ...(v)
Now,
$ar\left(\Delta ABC\right)=ar\left(\Delta AGB\right)+ar\left(\Delta BGC\right)+ar\left(\Delta AGC\right)$
$\Rightarrow ar\left(\Delta ABC\right)=ar\left(\Delta AGB\right)+ar\left(\Delta AGB\right)+ar\left(\Delta AGB\right)$ [From Eq. (v)]
$\Rightarrow ar\left(\Delta ABC\right)=3ar\left(\Delta AGB\right)$
$\Rightarrow ar\left(\Delta AGB\right)=\frac{1}{3}ar\left(\Delta ABC\right)$ ...(vi)
From Eqs. (v) and (vi)
$ar\left(\Delta BGC\right)=\frac{1}{3}ar\left(\Delta ABC\right)$
ad $ar\left(\Delta AGC\right)=\frac{1}{2}ar\left(\Delta ABC\right)$
Hence proved.