Question

Show that the diagonals of a parallelogram divide it into four triangles of equal area. [3 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Proof : 1 Mark

We know that diagonals of parallelogram bisect each other.
Therefore, O is the mid-point of AC and BD.
BO is the median in $\Delta ABC$. Therefore, it will divide it into two triangles of equal areas.

$\therefore ar\left(\Delta AOB\right)=ar\left(\Delta BOC\right)...\left(1\right)$

In $\Delta BCD,CO$ is the median.
$\therefore ar\left(\Delta BOC\right)=ar\left(\Delta COD\right)...\left(2\right)$

Similarly, $ar\left(\Delta COD\right)=ar\left(\Delta AOD\right)...\left(3\right)$

From equations (1), (2), and (3), we obtain
$ar\left(\Delta AOB\right)=ar\left(\Delta BOC\right)=ar\left(\Delta COD\right)=area\left(\Delta AOD\right)$
Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of equal area.