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\begin{document}

\small

\newpage

\section{补充梅涅劳斯定理}

\subsection{知识讲解}

\mybox{梅涅劳斯定理}

\begin{itemize}
\item \underlinebox{梅涅劳斯定理}
\begin{thbox}{梅涅劳斯定理}
如果一条直线与 $\triangle ABC$ 有三边 $BC,CA,AB$ 或其延长线交于 $D,E,F$ 三点,求证: $\dfrac{AF}{FB}\cdot\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}=1$.
\tcblower
\begin{prof}
\vspace{20em}
\end{prof}
\end{thbox}
\item \underlinebox{梅涅劳斯定理的逆定理}
\begin{thbox}{梅涅劳斯定理的逆定理}
若 $\triangle ABC$ 的三边 $BC,CA,AB$ 或其延长线上有三点 $D,E,F$, 且有 $\dfrac {AF}{FB}\cdot\dfrac {BD}{DC}\cdot\dfrac {CE}{EA}=1$, 那么 $D,E,F$ 三点共线.
\tcblower
\begin{prof}
\vspace{15em}
\end{prof}
\end{thbox}
\end{itemize}

\newpage

\subsection{典型例题}

\begin{eg}{}
$\triangle ABC$ 中, $D$ 是 $BC$ 的中点,经过 $D$ 的直线交 $AB$ 于 $E$, 交 $CA$ 延长线于 $F$. 求证: $\dfrac{FA}{FC}=\dfrac{EA}{EB}$.
\begin{center}\begin{tikzpicture}[scale=0.6]
\newpoint[][0]{0,3}AA[0]
\newpoint[][0]{-3.32,0}BB[-4]
\newpoint[][0]{2,0}CC[-4]
\newpoint[][0]{-0.66,0}DD[-4]
\newpoint[][0]{-1.21,4.82}EE[4]
\draw (E)--(C)--(B)--(A) (D)--(E);
\end{tikzpicture}\end{center}
\end{eg}

\vspace{20em}

\begin{eg}{}
如图, $\triangle ABC$ 中, $\angle ACB=90^{\circ}$, $AC=BC$. $AM$ 为 $BC$ 边上的中线, $CD\perp AM$ 于 $D$, $CD$ 延长线交 $AB$ 于 $E$. 求 $\dfrac{AE}{EB}$.
\begin{center}\begin{tikzpicture}[scale=0.7]
\newpoint[][0]{-3,0}AA[-4]
\newpoint[][0]{3,0}BB[-2]
\newpoint[][0]{0,3}CC[4]
\newpoint[][0]{0.6,1.2}DD[7]
\newpoint[][0]{1,0}EE[-4]
\newpoint[][0]{1.5,1.5}MM[1]
\draw (A)--(B)--(C)--(A) (C)--(E) (A)--(M);
\end{tikzpicture}\end{center}
\end{eg}

\vspace{20em}

\begin{eg}{}
若 $M$ 为 $\triangle ABC$ 内任一点,射线 $AM$, $BM$, $CM$ 分别交 $BC$, $CA$, $AB$ 于 $D$, $E$, $F$, 则有 $\dfrac{AM}{MD}=\dfrac{AF}{FB}+\dfrac{AE}{EC}$.
\begin{center}\begin{tikzpicture}[scale=0.7]
\newpoint[][0]{0,3.5}AA[4]
\newpoint[][0]{-3,0}BB[-4]
\newpoint[][0]{2,0}CC[-4]
\newpoint[][0]{-0.55,0}DD[-4]
\newpoint[][0]{0.9,1.93}EE[2]
\newpoint[][0]{-1.37,1.9}FF[6]
\newpoint[][0]{-0.34,1.32}MM[2.5][12pt]
\draw (A)--(B)--(C)--(A) (A)--(D) (B)--(E) (C)--(F);
\end{tikzpicture}\end{center}
\end{eg}

\vspace{20em}

\begin{eg}{}
在 $\triangle ABC$ 的三边 $BC$, $CA$, $AB$ 上分别取点 $D$, $E$, $F$, 使得 $\dfrac{BD}{DC}=\dfrac{CE}{EA}=\dfrac{AF}{FB}=\dfrac{1}{2}$. 若 $BE$ 与 $CF$, $CF$ 与 $AD$, $AD$ 与 $BE$ 的交点分别为 $A_1$, $B_1$, $C_1$. 求证: $\dfrac{S_{\triangle A_1B_1C_1}}{S_{\triangle ABC}}=\dfrac{1}{7}$.
\begin{center}\begin{tikzpicture}[scale=0.7]
\newpoint[][0]{0,4.4}AA[4]
\newpoint[][0]{-4,0}BB[-4]
\newpoint[][0]{1.28,0}CC[-4]
\newpoint[][0]{-2.24,0}DD[-4]
\newpoint[][0]{0.85,1.47}EE[1]
\newpoint[][0]{-1.33,2.93}FF[5]
\newpoint[][0]{0.16,1.26}G{A_1}[-4]
\newpoint[][0]{-0.96,2.51}H{B_1}[0]
\newpoint[][0]{-1.92,0.63}I{C_1}[5]
\draw (A)--(B)--(C)--(A) (A)--(D) (B)--(E) (C)--(F);
\end{tikzpicture}\end{center}
\end{eg}

\vspace{20em}

\begin{eg}{}
证明:三角形的外角平分线各与对边所在直线的交点,三点共线.
\begin{center}\begin{tikzpicture}[scale=0.45]
\newpoint[][0]{0,4}AA[4]
\newpoint[][0]{-5,0}BB[7]
\newpoint[][0]{-3,0}CC[-2]
\newpoint[][0]{4.13,0}DD[-2]
\newpoint[][0]{-4.36,-1.82}EE[-3]
\newpoint[][0]{-8.33,-2.67}FF[-4]
\draw (F)--(A)--(D)--(B)--(E)--(A) (C)--(F);
\draw[dashed,cyan] (F)--(D);
\end{tikzpicture}\end{center}
\end{eg}

\vspace{20em}

\begin{eg}{}
若 $\triangle ABC$ 与 $\triangle A_1B_1C_1$ 的对应顶点连线 $AA_1$, $BB_1$, $CC_1$ 相交于一点 $O$, 则对应边 $BC$ 与 $B_1C_1$, $CA$ 与 $C_1A_1$, $AB$ 与 $A_1B_1$ 的交点 $D$, $E$, $F$ 共线.
\begin{center}\begin{tikzpicture}[scale=0.35]
\newpoint[][0]{-1.83,0.63}AA[7]
\newpoint[][0]{2.13,0.05}BB[2]
\newpoint[][0]{0,1.7}CC[2]
\newpoint[][0]{-3.9,-4.73}R{A_1}[-5]
\newpoint[][0]{3.17,-2.54}S{B_1}[-2]
\newpoint[][0]{0,-5.59}T{C_1}[-2]
\newpoint[][0]{4.21,-1.54}DD[2]
\newpoint[][0]{-9.1,-3.59}EE[8]
\newpoint[][0]{8.52,-0.88}FF[0]
\newpoint[][0]{0,5.37}OO[4]
\draw (R)--(O)--(S) (E)--(C)--(D)--(T)--(E) (A)--(F)--(R) (O)--(T);
\draw[dashed,cyan] (F)--(E);
\end{tikzpicture}\end{center}
\end{eg}

\vspace{20em}

\subsection{课后习题}

\begin{enumerate}
\item
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\end{enumerate}

\subsection{习题参考答案}

\begin{enumerate}
\item
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\end{enumerate}




\end{document} 