Question

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

Answer 1

Here, $ABCD$ is a parallelogram with diagonals $AC$ and $BD.$

Diagonals of parallelogram bisect each other.

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

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

In $△ABC,$

Since, $OA=OC$ [ From ( 2 ) ]

$\therefore$ $BO$ is the median of $△ABC$

$\Rightarrow$ $ar\left(△AOB\right)=ar\left(△BOC\right)$ [ Median divides the triangle into equal area ] ---- ( 3 )

In $△ADC,$

Since, $OA=OC$ [ From ( 2 ) ]

$\therefore$ $DO$ is the median of $△ADC$

$\Rightarrow$ $ar\left(△AOD\right)=ar\left(△COD\right)$ [ Median divides the triangle into equal area ] ----- ( 4 )

Similarly, In $△ABD,$

Since $OB=OD$ [ From ( 1 ) ]

$\therefore$ $AO$ is the median of $△ABD$

$\Rightarrow$ $ar\left(△AOB\right)=ar\left(△AOD\right)$ [ Median divides the triangle onto equal area ] ---- ( 5 )

From ( 3 ), ( 4 ) and ( 5 )

$\Rightarrow$ $ar\left(△AOB\right)=ar\left(△BOC\right)=ar\left(△COD\right)=ar\left(△AOD\right)$ --- Hence proved

Hence, we have proved that the diagonals of a parallelogram divide into four triangles of equal area.