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

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#### Solution

We know that diagonals of parallelogram bisect each other.

Therefore, O is the mid-point of AC and BD.

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

∴ Area (ΔAOB) = Area (ΔBOC) ... (1)

In ΔBCD, CO is the median.

∴ Area (ΔBOC) = Area (ΔCOD) ... (2)

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

From equations (1), (2), and (3), we obtain

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

Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of equal area.

Concept: Corollary: Triangles on the same base and between the same parallels are equal in area.

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