prove that the diagonals of a parallelogram divides into four triangle of an 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
in triangle DCB = DEC=CEB ( as EC divides the triangle into 2 equal parts) equation 1
in triangle CBA= CEB=BEA ( as EB divides the triangle into two equal parts) equation 2
simillarly, in triangle BAD= AEB=BEA (as EA divides the triangle into two equal parts) equation 3
so by all the equation 1 2 3 we know that all the TRIANGLE ARE EQUAL
so DIAGONALS OF A PARALLELOGRAM DIVIDES IT INTO FOUR TRIANGLES OF EQUAL AREAS