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

To do:

We have to show that the diagonals of a parallelogram divide it into four triangles of equal area.

Solution:

Let in a parallelogram $ABCD$ diagonals $AC$ and $BD$ intersect at $O$.

We know that,

The diagonals of a parallelogram bisect each other.

This implies,

$OA = OC$ and $OB = OD$.

The median of a triangle divides it into two triangles of equal areas.

In $\triangle ABC$, $BO$ is the median.

This implies,

$ar (\triangle OAB) = ar (\triangle OBC)$.........(i)

In $\triangle ABD$, $AO$ is the median.

This implies,

$ar (\triangle OAB) = ar (\triangle OAD)$.......(ii)

In $\triangle ACD$, $DO$ is the median.

This implies,

$ar (\triangle OAD) = ar (\triangle OCD)$.........(iii)

From (i), (ii) and (iii), we get,

$ar (\triangle OAB) = ar (\triangle OBC) = ar (\triangle OCD) = ar(\triangle OAD)$

Hence proved.

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