Prove that the diagonals of a parallelogram divide it into four
triangles of equal areas therefore
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Let us consider in a parallelogram ABCD the diagonals AC and BD are cut at point O.
To prove: ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD)
Proof:
In parallelogram ABCD the diagonals bisect each other.
AO = OC
In ∆ACD, O is the mid-point of AC. DO is the median.
ar (∆AOD) = ar (COD) ….. (1) [Median of ∆ divides it into two triangles of equal arreas]
Similarly, in ∆ ABC
ar (∆AOB) = ar (∆COB) ….. (2)
In ∆ADB
ar (∆AOD) = ar (∆AOB) …. (3)
In ∆CDB
ar (∆COD) = ar (∆COB) …. (4)
From (1), (2), (3) and (4)
ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆AOD)
Hence proved.
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