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

Q : 3    Show that the diagonals of a parallelogram divide it into four triangles of equal area.

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Let ABCD is a parallelogram. So, AB || CD and AD || BC and we know that Diagonals bisects each other. Therefore, AO = OC and BO = OD

Since OD = BO
Therefore, ar ($\Delta$BOC) = ar ($\Delta$DOC)...........(a)   ( since OC is the median of triangle CBD)

Similarly, ar($\Delta$AOD) = ar($\Delta$DOC) ............(b)     ( since OD is the median of triangle ACD)

and, ar ($\Delta$AOB) = ar($\Delta$BOC)..............(c)           ( since OB  is the median of triangle ABC)

From eq (a), (b) and eq (c), we get

ar ($\Delta$BOC) = ar ($\Delta$DOC)= ar($\Delta$AOD) =  ($\Delta$AOB)

Thus, the diagonals of ||gm divide it into four equal triangles of equal area.

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