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A B C D is a parallelogram, G is the ...

`A B C D` is a parallelogram, `G` is the point on `A B` such that `A G=2 GB , E ` is a point of `D C` such that `CE=2DE` and `F` is the point of `BC` such that `BF=2F Cdot` Prove that: `ar EGB=1/6ar(ABCD)`

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Let the height of parallelogram ABCD and `triangleEGB` is the same.
Base of `triangleEGB=1/3AB`
area of `|\|`gm ABCD=`hxxAB`
ar(`triangleEGB`)=`1/2xx1/3ABxxh`
`1/6hxxAB`
`1/6xx`area of `|\|`gm ABCD
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