本帖最后由 青青子衿 于 2020-3-4 14:39 编辑
回复 3# kuing
不难求出
\begin{align*}
AD&=\frac{bc\sin A}{b\sin\frac{2A}3+c\sin\frac A3},\\
AE&=\frac{bc\sin A}{b\sin\frac A3+c\sin\frac{2A}3},
\end{align*}
kuing 发表于 2020-3-3 15:09
- \documentclass[tikz,border=4mm]{standalone}
- \usepackage{tkz-euclide}
- \usetkzobj{all}
- \usetikzlibrary{calc,shadows.blur}
- \begin{document}
- \begin{tikzpicture}
- \coordinate (A) at (5*0.2,5*1.1);
- \coordinate (B) at (0,0);
- \coordinate (C) at (5*1.7,0);
- \node [inner sep=2pt,label=left:$c$] (c) at ($ (A)!.5!(B) $) {};
- \node [inner sep=2pt,label=below:$b$] (b) at ($ (A)!.35!(C) $) {};
- \draw[thin,black,densely dashed](0,-1.2)--(B)(5*1.7,-1.2)--(C);
- \draw[thin,stealth-stealth](0,-0.7)--++(5*1.7,0)node[fill=white,midway]{$a$};
- \tkzFindAngle(B,A,C) \tkzGetAngle{at}
- \coordinate (D) at ($(A)!6cm!\at/3:(B)$);
- \tkzInterLL(A,D)(B,C)\tkzGetPoint{D}
- \draw (D) node[below]{$D$};
- \coordinate (E) at ($(A)!6cm!2*\at/3:(B)$);
- \tkzInterLL(A,E)(B,C)\tkzGetPoint{E}
- \draw (E) node[below]{$E$};
- \node [inner sep=1pt,label=left:$\color{red}{u}$] (p) at ($ (A)!.57!(D) $) {};
- \node [inner sep=1pt,label=left:$\color{blue}{v}$] (q) at ($ (A)!.55!(E) $) {};
- \begin{scope}[thin]
- \clip(B)--(A)--(D);
- \draw[fill=gray!20!white,semithick](A)circle(5mm);
- \draw[semithick](A)circle(5.8mm);
- \node (I) [label=-90:$\theta$,outer sep=15pt] at (A) {};
- \end{scope}
- \begin{scope}[thin]
- \clip(D)--(A)--(E);\draw[fill=gray!20!white,semithick](A)circle(5.4mm);
- \node (J) [label=-80:$\theta$,outer sep=13pt] at (A) {};
- \end{scope}
- \begin{scope}[thin]
- \clip(C)--(A)--(E);\draw[fill=gray!20!white,semithick](A)circle(5mm);
- \draw[semithick](A)circle(5.6mm);\draw[semithick](A)circle(6.2mm);
- \node (K) [label=-47:$\theta$,outer sep=9pt] at (A) {};
- \end{scope}
- \draw[thick,red] (A)--(D);
- \draw[thick,blue] (A)--(E);
- \draw[thick] (A) node[above]{$A$}--
- (B) node[left]{$B$}--
- (C) node[right]{$C$}--(A);
- \node[fill=yellow!80,blur shadow={shadow xshift=-0.5ex},
- text width=20em,anchor=south west,rounded corners] at
- ([xshift=3em]b.east)
- {\[{\color{magenta}{16S^2}}=(a+b+c)(a+b-c)(a+c-b)(b+c-a)\]
- \[\frac{a^2b^2-(b^2-c^2)^2}{bc^2}\textcolor{red}{u}^3+3a^2\textcolor{red}{u}^2-{\color{magenta}{16S^2}}=0\]
- \[\frac{a^2c^2-(b^2-c^2)^2}{b^2c}\textcolor{blue}{v}^3+3a^2\textcolor{blue}{v}^2-{\color{magenta}{16S^2}}=0\]
- ~};
- \end{tikzpicture}
- \end{document}
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