Data: Diagonals of 
ABCD, AC and BD intersect at ‘O’.
To Prove: ar.(∆AOD) = ar.(∆DOC) = ar.(∆COB) = ar.(∆AOB)
Proof: ‘O’ is the mid-point of diagonals AC and BD. DO is the median of ∆ADC.
∴ ar.(∆AQD) = ar.(∆DOC) ……….. (i)
CO is the median of ∆DCB.
∴ ar. (∆DOC) = ar.(∆BOC) ……….. (ii)
BO is the median of ∆ABC.
∴ ar.(∆BOC) = ar.(∆AOB) …………. (iii)
AO is the median of ∆ADB.
∴ ar.(∆AOB) = ar.(∆AOD) …………. (iv)
From (i), (ii), (iii) and (iv),
ar. (∆AOD) = ar. (∆DOC) = ar. (∆COB) = ar. (∆AOB)
∴ The diagonals of a parallelogram divide it into four triangles of equal area.
