<s>[INST] Trisect a quadrilateral into a $9$-grid; the middle has $1/9$ the area Trisect sides of a quadrilateral and connect the points to have nine quadrilaterals, as can be seen in the figure. Prove that the middle quadrilateral area is one ninth of the whole area.
 [/INST] This is most easily done using vectors. Let the points $A, B, C, D$ be represented by the vectors $a, b, c, d$. The area $[ABCD]$ is equal to $\frac{1}{2}(a-c) \times (b-d) $. 
If you are unfamiliar with this, consider triangulation using the origin, and sum up the 4 triangle areas, to get
$$\begin{align} [ABCD] = & \frac{1}{2} a \times b + \frac{1}{2} b \times c + \frac{1}{2} c \times d + \frac{1}{2} d \times a \\ = & (a-c) \times \frac{1}{2} b + (a-c ) \times (-\frac{1}{2} d) \\= &  \frac{1}{2}(a-c) \times (b-d) \end{align}$$
It is easy to show that $W= \frac{4a+2b+c+2d} {9}, X = \frac{ 2a+4b+2c+d} { 9},Y = \frac{a+2b+4c+d} { 9} , Z = \frac{ 2a+b+2c+4d} {9}  $. Hence the area is
$$
[WXYZ] = \frac{1}{2} ( \frac{3a-3c}{9} ) \times ( \frac{3b-3d}{9} ) = \frac{1}{9} \times \frac{1}{2} (a-c)(b-d) = \frac{1}{9} [ABCD]$$
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