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D在BC上, E在AB上, F为AD与BC的交点, 则
1area (△ AEC) + 1area (△ ADC) = 1area (△ AFC) + 1area (△ ABC).
来源Wikipedia
\begin{tikzpicture}[scale=0.5,inner sep=.2]
\node (v1) at (-6.3,-4.6) {A};
\node (v2) at (1.1,0.6) {B};
\node (v3) at (2.7,-4.6) {C};
\node (v4) at (-2.5,-2) {E};
\node (v5) at (2,-2.5) {D};
\node (v6) at (-0.32,-3.07) {F};
\draw (v1)edge(v6) --(v4)-- (v2) --(v5)-- (v3) edge (v6)-- (v1)(v5)--(v6)--(v4);
\end{tikzpicture}证明\[{\S{AFC}\over\S{AEC}}+{\S{AFC}\over\S{ADC}}={\S{AFBC}\over\S{ABC}}+{\S{ABFC}\over\S{ABC}}=1+{\S{AFC}\over\S{ABC}}\]
当$AE\px CD$时, $\frac1{\S{ABC}}=0$, 得到crossed ladders theorem |
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