`A B C D` is a parallelogram, `G` is the point on `A B` such that `A G=2G B ,E` is a point of `D C` such that `C E=2D E` and `F` is the point of `B C` such that `B F=2F Cdot` Prove that : `a r(A D E G)=a r(G B C E)` `a r( E G B)=1/6a r(A B C D)`

Explanation

`AG=2GB, " EC=2DE"`

`AB=CD`

`ED=1/3CD" ED="1/3 AB`

`EC=2/3CD" EC="2/3AB`

`AG=2/3AB" BG="1/3AB`

`AG=EC" BG="ED`

`"Hence" ar(ADEG)=ar(GBCE)->"(due to having same parallel sides and height )"`

`ar(/_\EGB)=1/2xx"base"xx"height"`

...