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