Question:
ABCD is a parallelogram whose diagonals intersect at O .If P is any point on BO, prove that:
(i) ar(ΔADO) = ar(ΔCDO).
(ii) ar(ΔABP) = 2ar(ΔCBP).
Solution:
Given that ABCD is the parallelogram
To Prove:
(i) ar(ΔADO) = ar(ΔCDO).
(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.
Then ar(ΔADO) = ar(ΔCDO).
(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).
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