`A B C D` is a parallelogram, `G` is the point on `A B` such that `A G=2 G B` ,` E ` is a point of `D C` such that `C E=2DE` and `F` is the point of `BC` such that `B F=2F Cdot` Prove that: Find what portion of the area of parallelogram is the area of ` `EFG`

Explanation

ar(`triangleEFG`)=ar(trap. BGEC)-ar(`triangleBGF`)

Now, ar(trap. BGEC)=`1/2(GB+EC)xxEP`

`1/2(1/3AB+2/3CD)xxEP`

`1/2ABxxEP`

`1/2ar(|\|gm ABCD)`

ar(`triangleBGF`)=`1/2xxBFxxGR`

`1/2xx2/3BCxxGR`

`2/3xx1/2BCxxGR`

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