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

(N/A) Let $ABCD$ be a parallelogram such that its diagonals $AC$ and $BD$ intersect at $O$.
Since the diagonals of a parallelogram bisect each other,we have $AO = OC$ and $BO = OD$.
Draw $CE \perp BD$.
Now,$\text{ar}(\Delta BOC) = \frac{1}{2} \times BO \times CE$ and $\text{ar}(\Delta DOC) = \frac{1}{2} \times OD \times CE$.
Since $BO = OD$,it follows that $\text{ar}(\Delta BOC) = \text{ar}(\Delta DOC) \quad ... (1)$.
Similarly,by drawing perpendiculars from $A$ to $BD$,we can show that $\text{ar}(\Delta AOD) = \text{ar}(\Delta AOB) \quad ... (2)$.
Also,considering $\Delta ABD$ and $\Delta CBD$ on the same base $BD$ and between the same parallels $AD \parallel BC$,we know that $\text{ar}(\Delta ABD) = \text{ar}(\Delta CBD)$.
Since $\text{ar}(\Delta ABD) = \text{ar}(\Delta AOB) + \text{ar}(\Delta AOD)$ and $\text{ar}(\Delta CBD) = \text{ar}(\Delta BOC) + \text{ar}(\Delta DOC)$,we have $\text{ar}(\Delta AOB) + \text{ar}(\Delta AOD) = \text{ar}(\Delta BOC) + \text{ar}(\Delta DOC)$.
Using the relations derived,all four triangles have equal area: $\text{ar}(\Delta AOB) = \text{ar}(\Delta BOC) = \text{ar}(\Delta COD) = \text{ar}(\Delta DOA)$.
Thus,the diagonals of a parallelogram divide it into four triangles of equal area.