How do you prove $S_{XYZT} \leq \dfrac{1}{5} S_{ABCD} $? In the given figure, $ABCD$ is a convex quadrilateral. Suppose that $M, N, P, Q$ are mid-points of $AB, BC, CD, DA$, respectively. Prove that $S_{XYZT} \leq \dfrac{1}{5} S_{ABCD} $ where $S_{ABCD}$ (resp. $S_{XYZT}$) is the area of $ABCD$ (resp. $XYZT$)? 

Could you please give a key hint to solve this exercise? Thank you so much for your discussions!
 A: As already pointed out in a comment the ratio
$\dfrac{S_{ABCD}}{S_{TXYZ}}$ is not exactly equal to 5, though it is surprisingly close to the value almost in any convex quadrilateral. Only if one side tends to 0 (so that the quadrilateral degenerates to triangle) the ratio tends to 6 (as it also should).
There is however a class of quadrilaterals for which the ratio is exactly 5. This class is parallelograms, and the proof in this case is simple.

As easy to understand for any convex quadrilateral:
$$
S_{TXYZ}=S_{AXM}+S_{BYN}+S_{CZP}+S_{DTQ}.
$$
and
$$
S_{ABCD}-S_{TXYZ}=S_{AYB}+S_{BZC}+S_{CTD}+S_{DXA}.
$$
Specifically for parallelogram we have:
$$
S_{AYB}=4S_{AXM}, \dots
$$
Thus,
$$S_{ABCD}-S_{TXYZ}=4S_{TXYZ}.$$

UPDATE:
On the basis of numerical evidence I would conjecture the following statement:
For any convex quadrilateral $$5\le\dfrac{S_{ABCD}}{S_{TXYZ}}<6$$ and the ratio is equal to 5 if and only if the quadrilateral $TXYZ$ is a trapezoid.
For the characterization of the quadrilateral $ABCD$ the above statement means that its vertices lie on four equidistant parallel lines, two opposite vertices being on the external lines (see figure below). I do not know if a special name for such a quadrilateral exists.
To prove the "if" part of the statement only a slight modification of the previous proof (for parallelogram) is required due to the fact that $S_{AXM}=S_{DTQ}$ and $S_{BYN}=S_{CZP}$.

A: ►Let the four vertex $C=(0,0),D=(d_1,d_2),A=(a_1,a_2),B=(b_1,b_2)$.
►Calculation determines lines $\overline{BQ},\overline{ND},\overline{MC},\overline{AP}$.
►Points $Y=\overline{BQ}\cap\overline{MC}\\X=\overline{BQ}\cap\overline{AP}\\Z=\overline{ND}\cap\overline{MC}\\T=\overline{ND}\cap\overline{AP}$
►Do you know how to calculate directly the area of a convex quadrilateral? For example for $CDAB$ put the coordinates as follows starting from an arbitrary vertice and contrary to clockwise direction, say starting with $C=(0,0)$
$$0\hspace{10mm}0 \\d_1\hspace{10mm}d_2\\a_1\hspace{10mm}a_2\\b_1\hspace{10mm}b_2\\0\hspace{10mm}0$$ You must finish repeating the first chosen vertice.Then here you have the area is given for 
$$\frac12[(0\cdot d_2+d_1\cdot a_2+a_1\cdot b_2+b_1\cdot0)-(0\cdot  b_2+b_1\cdot a_2+a_1\cdot d_2+d_1\cdot0)]$$ (Multiplication descending for positive parenthesis and ascending for the negative one).
Repeat this with the smaller quadrilateral and compare.
