# 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

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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

HENCE PROVED

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