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[几何] 四边形 线段比

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hbghlyj Posted 2022-3-12 22:42 |Read mode
S A B C D E F G H设四边形ABCD相对的两组顶点A,C和B,D与不在四边形的边或它们的延长线上的一点S联结而成的四条直线,与对角线BD和AC或它们延长线依次交于点E,G和F,H,则有$$\frac{A F}{F C} \cdot \frac{B G}{G D} \cdot \frac{C H}{H A} \cdot \frac{D E}{E B}=1$$


证明\begin{align*}&\frac{A F}{F C} \cdot \frac{B G}{G D} \cdot \frac{C H}{H A} \cdot \frac{D E}{E B}\\ =&\frac{\S{ABS}}{\S{BCS}}·\frac{\S{BCS}}{\S{CDS}}·\frac{\S{CDS}}{\S{DAS}}·\frac{\S{DAS}}{\S{ABS}}\\=&1\end{align*}

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 Author| hbghlyj Posted 2022-3-14 03:04
或者可以通过交比相等
(AC;FH)=(EG;BD)

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