<s>[INST] The area of interior quadrilateral formed by connecting trisect points and vertices of larger quadrilateral I drew a figure on GeoGebra to explore the area of the smaller quadrilateral formed by joining the vertices of the larger quadrilateral and the trisect points on the edges of the larger quadrilateral.
Here is what I think to be correct:
$$Area_{\Delta E_2H_2G_2F_2}=\frac{2}{5}Area_{\Delta A_1B_1C_1D_1}$$
I haven't figured out a way to prove this question yet. If this is correct, could any of you provide a possible solution to the problem?

 [/INST] Some example caculation for retangular case. Let $ABCD$ be a rectangular with $AB=6,AD=3$.
Let $TQ=x$,$XM=y$ and $\angle ZTX=\alpha$, then $XM=ZP=x$; $TQ=YN=y$; $DX=BZ=3x$;$AT=CY=3y$ and $XYZT$ is parallelogram rhombus with $XY=6x$ and $XT=6y$
We have $37a^2=(6a)^2+a^2=AB^2+AQ^2=BQ^2=100x^2$, then $x=\frac{\sqrt{37}}{10}a$
$13a^2=(2a)^2+(3a)^2=AD^2+DM^2=AM^2=(10y)^2=100y^2$, then $y=\frac{\sqrt{13}}{10}a$
and $cos(\alpha)=\frac{TQ^2+TA^2-AQ^2}{2TQ.TA}=\frac{x^2+9y^2-a^2}{6xy}=\frac{9}{\sqrt{481}}$, so $sin(\alpha)=\frac{20}{\sqrt{481}}$
Then $S_{XYZT}=sin(\alpha)XY.XT=\frac{20}{\sqrt{481}}.36\frac{\sqrt{481}a^2}{100}=7.2a^2$, and $S_{ABCD}=18a^2$ so $\frac{S_{XYZT}}{S_{ABCD}}=\frac{2}{5}$ 
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