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Question

In the figure, diagonals AC and BD of a quadrilateral ABCD intersect at O such that OB=OD, show that ar(ΔDOC)=ar(ΔAOB)
1151028_927ff572f3e349cfaef6aa0ddaca0183.png

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Solution

BO=OD
Since BO=OD is the mid point
we know thta a mid point of a divides it into two of equal areas
CO is medpoint of BCD
i.e., ar(COD)=ar(COB)(1)
AO is a medpoint of ABD
i.e. ar(AOD)=ar(AOB)(2)
from (I) & (II) we have
ar(COD)+ar(AOD)=ar(COB)+ar(AOB)
ar(ADC)=ar(ABC)


1376941_1151028_ans_91d1e0f8315a44419ed9f3fc8751b5dd.png

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