Show that the diagonals of a parallelogram divide it into four triangles of equal area.
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.
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- Corollary: Triangles on the same base and between the same parallels are equal in area.