ABCD is a parallelogram whose diagonals intersect at O. IF P is any point on BO, prove that
(i) $a r (△ A D O) = a r (△ C D O) (i i) a r (△ A B P) = a r (△ C B P)$

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Solution

Given : In || gm ABD, Diagonals AC and BD intersect each other at O
P is any point on BO
AP and CP are Joined

To prove:
(i) ar $(△ A D O) = a r (△ C D O) (i i) a r (△ A B P) = a r (△ C B P)$
Proof:
(i) in $△ A D C,$
O is the mid point of AC
$∴ a r (△ A D O) = a r (△ C D O)$
(ii) Since O is the mid point of AC
$∴ P O i s t h e m e d i a n o f △ A P C ∴ a r (△ A P O = a r (△ C P O)$
Similarly, BO is the median of $△ A B C$
$∴ a r (△ A B O) = a r (△ B C O)$
Subtracting (i) from (ii),
ar $(△ A B O) - a r (△ A P O) = a r (△ B C O) - a r (△ C P O) \Rightarrow a r (△ A B P) = a r (△ C B P)$

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ABCD is a parallelogram whose diagonals intersect at O. IF P is any point on BO, prove that (i) ar(△ADO)=ar(△CDO)(ii)ar(△ABP)=ar(△CBP)

ABCD is a parallelogram whose diagonals intersect at O. IF P is any point on BO, prove that
(i) $a r (△ A D O) = a r (△ C D O) (i i) a r (△ A B P) = a r (△ C B P)$