ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that

(i) ar(Δ ADO) = ar(Δ CDO)

(ii) ar(Δ ABP) = ar(Δ CBP)

Given that,

ABCD is a parallelogram

To prove:

(i) Area () = Area ()

(ii) Area () = Area ()

Proof: We know that,

Diagonals of a parallelogram bisect each other

Therefore,

AO = OC and,

BO = OD

(i) In ΔDAC, DO is a median.

Therefore,

Area (ΔADO) = Area (ΔCDO)

Hence, proved

(ii) In , since BO is a median

Then,

Area (ΔBAO) = Area (ΔBCO) (i)

In a ΔPAC, since PO is the median

Then,

Area (ΔPAO) = Area (ΔPCO) (ii)

Subtract (ii) from (i), we get

Area (ΔBAO) - Area (ΔPAO) = Area (ΔBCO) - Area (ΔPCO)

Area (ΔABP) = Area (ΔCBP)

Hence, proved