Question

Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Answer 1

According to the question,

Given: A parallelogram $ABCD$ with diagonals $AC$ & $BD$

Proof: $ABCD$ is a parallelogram Diagonals of a parallelogram bisect each other

$O$ is the mid-point of $BD$, i.e, $OB=OD$ ........(1)

& $O$ is the mid-point of $AC$, i.e., $OA=OC$ ........(2)

In triangle $ABC$,

Since $OA=OC$ From (2)

$BO$ is the median of triangle $ABC$

implies ar $\left(\Delta AOB\right)$ $=$ ar $\left(\Delta BOC\right)$