Q. 73.8( 10 Votes )

*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*)

Answer :

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 (Δ*ADO*) = Area (Δ*CDO*)

Hence, proved

**(ii)** In , since B*O* is a median

Then,

Area (Δ*BAO*) = Area (Δ*BCO*) (i)

In a ΔPAC, since PO is the median

Then,

Area (ΔPAO) = Area (ΔPCO) (ii)

Subtract (ii) from (i), we get

Area (Δ*BAO*) - Area (ΔPAO) = Area (Δ*BCO*) - Area (ΔPCO)

Area (Δ*ABP*) = Area (Δ*CBP*)

Hence, proved

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