Question

Show that:
A diagonal divides a parallelogram into two triangles of equal area.

Answer 1

Here, $ABCD$ is a parallelogram and $BD$ is diagonal.

In $△ADB$ and $△CBD$

$AD\parallel BC$

$\Rightarrow$ $\angle ADB=\angle CBD$ [ Alternate angles ]

Also $AB\parallel DC$

$\Rightarrow$ $\angle ABD=\angle CDB$ [ Alternate angles ]

$\Rightarrow$ $DB=BD$ [ Common side ]

$\therefore$ $△ADB\cong △CBD$ [ By $SAS$ congruence rule ]

Since congruent figures have same area.

$\therefore$ $ar\left(ADB\right)=ar\left(CBD\right)$ [ Hence, proved ]

$\therefore$ We have proved that a diagonal divides a parallelogram into two triangles of equal area.