[原创]平行线分线段成比例推广

如图, 在 $\triangle ABC$ 中, 点 $E,F$ 分别为边 $AC,AB$ 上的点, 且 $BE$ 与 $CF$ 交于 $G$ 点, 连接 $AG$ 并延长交 $BC$ 于点 $D$. 可以得到
\[\frac{GE}{GB} = \frac{AF}{FB} \cdot \frac{AF}{FB},~\frac{GF}{GC} = \frac{AE}{EC} \cdot \frac{BF}{AB}.\]
进一步可以得到
\[\frac{DB}{DC}=\frac{\, BF/AF \,}{\, CE/AE \,}, \frac{AG}{GD}=\frac{AF}{BF}+\frac{AE}{CE}, \]
即 Ceva 定理
\[\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1,\]
及其扩展
\[ \frac{BG}{GE} \cdot \frac{EC}{CA} \cdot \frac{AF}{FB} = 1, \frac{CG}{GF} \cdot \frac{FB}{BA} \cdot \frac{AE}{EC} = 1. \]

证 过 $E$ 作 $CF$ 的平行线交 $AB$ 于点 $H$.
\[ \frac{BG}{GE} = \frac{BF}{FH} = \frac{BF}{FA \cdot EC/CA} = \frac{BF}{FA} \cdot \frac{AC}{CE}, \]
即
\[ \frac{BG}{GE} \cdot \frac{EC}{CA} \cdot \frac{AF}{FB} = 1. \]
同理
\[ \frac{CG}{GF} \cdot \frac{FB}{BA} \cdot \frac{AE}{EC} = 1. \]

注意到
\[ \frac{BG}{GE} \cdot \frac{EA}{AC} \cdot \frac{CD}{DB} = 1, \]
则
\[ \frac{EC}{CA} \cdot \frac{AF}{FB} = \frac{EA}{AC} \cdot \frac{CD}{DB}, \]
即 Ceva 定理
\[ \frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1. \]

另外同理可得
\[\frac{AG}{GD} \cdot \frac{DB}{BC} \cdot \frac{CE}{EA} = 1 \]
\[\frac{AG}{GD} = \frac{CB}{BD} \cdot \frac{AE}{EC} = \frac{CD}{DB} \cdot \frac{AE}{EC} + \frac{AE}{EC}. \]
利用 Ceva 定理即得
\[\frac{AG}{GD} = \frac{AF}{FB} + \frac{AE}{EC}.\]

一个简单例子: 若 $E,F$ 均为中点, 即 $AE/EC=AF/FB=1$, 则 $BD/DC=1$ 且 $AG/GD=1+1=2$, 也就是 $D$ 为中点, 且三条中线交于一点, 重心 $G$ 的位置也确定.

顺便附上该图的 $\LaTeX$ 代码:

\documentclass{standalone}

\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\begin{document}

\begin{tikzpicture}[very thick, scale=2.04, transform shape]

\coordinate [label=above:$A$] (A) at (3,3);
\coordinate [label=left:$B$] (B) at (0,0);
\coordinate [label=right:$C$] (C) at (4,0);

\draw [name path = CF] ($(A)!2/3!(B)$) coordinate [label=left: $F$] (F) -- (C);
\draw [name path = BE] ($(A)!1/2!(C)$) coordinate [label=right: $E$] (E) -- (B);
\path [name intersections = {of = CF and BE,by = { [ label=below: $G$ ] G } }]; %% {of 之间不能有空格
\draw ($(C)!2/3!(B)$) coordinate [label=below: $D$] -- (A)
		($(A)!1/2!(F)$) coordinate [label=left: $H$] -- (E)
		(A) -- (B) -- (C) -- cycle;

\node [left] at ($(B)!1/6!(A)$) {$x$};
\node [below] at ($(B)!1/6!(C)$) {$z$};
\node [right] at ($(C)!1/4!(A)$) {$y$};

\end{tikzpicture}

\end{document}