In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.

If BO = OD, prove that


Ar(∆ABC) = ar(∆ADC).


Given : A quadrilateral ABCD with diagonals AC and BD and BO = OD


To prove: Area of (∆ABC) = area of (∆ADC)


Proof : BO = OD [given]


Here AO is the median of ∆ABD


Area of (∆AOD) = Area of (∆AOB) ---------------- 1


And OC is the median of ∆BCD


Area of (∆COD) = Area of (∆BOC) ---------------- 2


Now by adding –1 and –2 we get


Area of (∆AOD) + Area of (∆COD) = Area of (∆AOB) + Area of (∆BOC)


Area of (∆ABC) = Area of (∆ADC)


Hence proved


16