The given figure shows a pentagon ABCDE.EG drawn parallel to DA meets BA produced at G and CF drawn parallel to DB meets AB produced at F. Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.

Explanation

`ar(triangleDBC)=ar(triangleDBF)`
`ar(triangleADE)=ar(triangleDBF)`
`ar(triangleDBC)+ar(triangleADE)+ar(triangleABD)=ar(triangleDBF)+ar(triangleADG)+ar(triangleABD)`
`therefore ar("pentagon"ABCDE)=ar(triangleDGF)`.