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

Question:

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

(ii) ar(ΔABP) = 2ar(ΔCBP).

Solution:

Given that ABCD is the parallelogram

To Prove:

(ii) ar(ΔABP) = 2ar(ΔCBP).

Proof:

we know that diagonals of parallelogram bisect each other

∴  AO = OC and BO = OD

(i) In ΔDAC, since DO is a median.

(ii) In ΔBAC, since BO is a median.

Then ar(ΔBAO) = ar(ΔBCO) ⋅⋅⋅⋅ (1)

In ΔPAC, since PO is a median.

Then ar(ΔPAO) = ar(ΔPCO) ⋅⋅⋅⋅ (2)

Subtract equation 2 from 1.

⇒ ar(ΔBAO) − ar(ΔPAO) = ar(ΔBCO) − ar(ΔPCO)

⇒ ar(ΔABP) = 2ar(ΔCBP).