四边形 九宫格 中间的面积为1/9

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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.

math.stackexchange.com/questions/465986/trisect-a-quadrilateral- ... dle-has-1-9-the-area

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

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相关问题:jwilson.coe.uga.edu/EMAT6680Fa08/Kuzle/Essay%202/Quadrilaterals%20Revealed.html