Geometry Question about Area and surface

Question

Problem.

According to following diagram, prove (Area of (MM'N'N)) = 1/3*(Area of ABCD)). We Know that AN = NM = MB and DN' = N'M' = M'C. and quadrilateral ABCD is not and special quadrilateral. diagram: http://uploads.im/Jh2MF.png

Answer 1

edited proof:

As we might observe $S_{ACM}=\dfrac23S_{ACB}$ and $S_{ACN'}=\dfrac23S_{ACD}$, so in total $$S_{AMCN'}=\frac23S_{ABCD}$$ Now consider $\triangle MCN'$, since $MM'$ is median $S_{MN'M'}=S_{MM'C}$.

Likewise we can conclude $S_{NN'M}=S_{AN'N}$

By adding these two we have: $$S_{NN'M}+S_{MN'M'}=S_{AN'N}+S_{MM'C}\Rightarrow S_{MM'N'N}=S_{AN'N}+S_{MM'C}\\ \Rightarrow S_{MM'N'N}=\frac12 S_{AMCN'}=\frac12\cdot\frac23S_{ABCD}\quad\checkmark$$