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

We know that the diagonals of a parallelogram bisect each other.

Also, the median of a triangle divides it into two triangle of equal areas. By the use of these observations, we can get the required result.

Let's draw a diagram according to the question statement. 

We know that diagonals of parallelograms bisect each other. Therefore, O is the mid-point of diagonal AC and BD.

BO is the median in ΔABC. Therefore, BO will divide ΔABC into two triangles of equal areas

∴ Area (ΔAOB) = Area (ΔBOC) ... (1)

Also, In ΔBCD, CO is the median. Therefore, median CO will divide ΔBCD into two equal triangles.

Hence, Area (ΔBOC) = Area (ΔCOD) ... (2)

Similarly, Area (ΔCOD) = Area (ΔAOD) ... (3)

From Equations equation (1), (2), and (3) we obtain,

Area (ΔAOB) = Area (ΔBOC) = Area (ΔCOD) = Area (ΔAOD)

Therefore, we can say that the diagonals of a parallelogram divide it into four triangles of equal area.