TITLE: Trisect a quadrilateral into a $9$-grid; the middle has $1/9$ the area
QUESTION [15 upvotes]: 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.

REPLY [6 votes]: Consider all occurring points as vectors, as in @Calvin Lin's answer, and write $\mu$ for ${1\over3}$. Then
$$p=(1-\mu)a+\mu b,\quad h=(1-\mu)d+\mu c,\quad n=(1-\mu)a+\mu d,\quad e=(1-\mu) b+\mu c\ .$$
It follows that
$$(1-\mu)p+\mu h=(1-\mu)n+\mu e\quad(=:w')\ ,$$
which shows that in fact 
$$w=w'=(1-\mu)^2 a +\mu(1-\mu)(b+d)+\mu^2 c\ .$$
Interchanging $a$ and $c$ here gives
$$y=(1-\mu)^2 c +\mu(1-\mu)(b+d)+\mu^2 a\ ,$$
so that we arrive at
$$w-y=(1-2\mu)(a-c)\ .$$ Appealing to symmetry again we conclude that we also have
$$x-z=(1-2\mu)(b-d)\ .$$
It follows that  $${\rm area}[WXYZ]=(1-2\mu)^2\ {\rm area}[ABCD]\ ,$$ 
and this holds for any $\mu\in[0,{1\over2}[\ $.

REPLY [4 votes]: Claim. $W$ and $Z$ trisect $\overline{PH}$. Likewise elsewhere.
Proof. Left to the reader (for now).

Given the claim, we can make this illustrated argument:

Here, we have $\triangle ABC \sim \triangle PBF$, with
$$\frac{|\overline{PB}|}{|\overline{AB}|} = \frac{|\overline{FB}|}{|\overline{CB}|} = \frac{2}{3} = \frac{|\overline{PF}|}{|\overline{AC}|} 
\qquad \text{and} \qquad
\overline{PF} \parallel \overline{AC}$$
and $\triangle PZF \sim \triangle WZY$, with
$$\frac{|\overline{WZ}|}{|\overline{PZ}|} = \frac{|\overline{YZ}|}{|\overline{FZ}|} = \frac{1}{2} = \frac{|\overline{WY}|}{|\overline{PF}|} 
\qquad \text{and} \qquad
\overline{WY} \parallel \overline{PF}$$
so that
$$|\overline{WY}|= \frac13 |\overline{AC}| \qquad \text{and} \qquad \overline{WY} \parallel \overline{AC}$$
and likewise
$$|\overline{XY}|= \frac13 |\overline{BD}| \qquad \text{and} \qquad \overline{XZ} \parallel \overline{BD}$$
By the Diagonal-Diagonal-Angle formula for quadrilateral area,
$$|\square WXYZ| = \frac{1}{2}|\overline{WY}||\overline{XZ}|\sin\theta = \frac12 \cdot \frac{1}{3}|\overline{AC}| \cdot \frac13 |\overline{BD}|\cdot \sin\theta = \frac19 |\square ABCD|$$