|
|
|
Last edited by hbghlyj at 2025-3-19 21:13:56D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
%20at%20(-6.3,-4.6)%20%7BA%7D;%0A%5Cnode%20(v2)%20at%20(1.1,0.6)%20%7BB%7D;%0A%5Cnode%20(v3)%20at%20(2.7,-4.6)%20%7BC%7D;%0A%5Cnode%20(v4)%20at%20(-2.5,-2)%20%7BE%7D;%0A%5Cnode%20(v5)%20at%20(2,-2.5)%20%7BD%7D;%0A%5Cnode%20(v6)%20at%20(-0.32,-3.07)%20%7BF%7D;%0A%5Cdraw%20(v1)edge(v6)%20--(v4)--%20(v2)%20--(v5)--%20(v3)%20edge%20(v6)--%20(v1)(v5)--(v6)--(v4);%0A%5Cend%7Btikzpicture%7D) 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 |
|
