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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.enter image description here

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  • $\begingroup$ Very nice picture. I would start by separating quadrilaterals having one or two pairs of parallel sides from those with no parallel sides, then taking the non-parallel sides and extending them to where they meet, then applying triangle methods to what remains. $\endgroup$ – abiessu Aug 12 '13 at 17:26
  • $\begingroup$ It was a homework when I was a high school student. I have just recalled it. $\endgroup$ – Amir Kazemi Aug 12 '13 at 17:29
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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}[\ $.

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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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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:

Figures

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, $$|\square WXYZ| = \frac{1}{2}|\overline{WY}||\overline{XZ}|\sin\theta = \frac12 \cdot \frac{1}{3}|\overline{AC}| \cdot \frac13 |\overline{BD}|\cdot \sin\theta = \frac19 |\square ABCD|$$

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Stretch the figure in a direction and by an amount which makes the top and bottom edges parallel. Such a transformation preserves relative areas. Now look at each trapezoid in the center row. It is clear that the area is equal to the average of the areas of the trapezoids above and below. Similarly, stretch to make the sides parallel, and look at the trapezoids in the center column. The area of each is equal to the average of the areas of the figures to the right and left. It follows that the area of the center quadrilateral is equal to the average of the areas of the outer eight quadrilaterals.

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    $\begingroup$ You cannot make nonparallel lines parallel by means of an affine stretching. $\endgroup$ – Christian Blatter Aug 13 '13 at 10:34
  • $\begingroup$ Yes, I see. Because that would require sending the point of intersection to infinity. Thank you. $\endgroup$ – Bill Kleinhans Aug 13 '13 at 21:47

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