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


14