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In a quadrilateral ABCD, it is being giv...

In a quadrilateral ABCD, it is being given that M is the midpoint of AC. Prove that
`ar(squareABMD)=ar(squareDMBC)`.

Text Solution

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Median BM divides `triangle ABC` into two `triangle` of equal area.
`therefore ar(triangleABM)=ar(triangleDMC)`.
Median DM divides `triangle`DAC into two `triangle` of equal area.
`therefore ar(triangleAMD)=ar(triangleDMC)." "...(ii)`
From (i) and (ii) , we get
`ar(triangleABM)+ar(triangleAMD)=ar(triangleBMC)+ar(triangleDMC)`
`rArr ar(squareABMD)=ar(squareDMBC)`.
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