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

#### Complete Python Prime Pack

9 Courses     2 eBooks

#### Artificial Intelligence & Machine Learning Prime Pack

6 Courses     1 eBooks

#### Java Prime Pack

9 Courses     2 eBooks

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.

Updated on 10-Oct-2022 13:41:50