ABCD is a parallelogram. E is a point on BA such that `BE =2 EA` and F is a point on Do such that DF=2 FC. Prove that AE CF is a parallelogram whose area is one third area of parallelogram `ABCD`

Explanation

Given: `BE=2EA` and `DF=2FC`
To Prove: `ar(∥^(gm) AECF)=1/3 ar(∥^(gm) ABCD)`
Proof: `BE=2EA` and `DF=2FC`
`=>AB−AE=2AE` and `DC−FC=2FC`
`=>AB=3 AE` and `DC=3FC`
`=>AE=1/3AB` and `FC=1/3DC`
`=>AE=FC` [`AB=DC`]
`:.AEFC` is a parallelogram.
Clearly, parallelogram `ABCD` and `AEFC` have the same altitude and `AE=1/3AB`.
`ar(∥^(gm) AECF)=hxx1/3AB=1/3xxhxxAB=1/3ar(∥^(gm) ABCD)`
`:.ar(∥^(gm) AECF)=1/3 ar(∥^(gm) ABCD)`
Hence Proved.