如何在 LaTeX 中绘制几何三角形？

Question 1

使用纯的tikz，使用极坐标和交点：

\documentclass[11pt, tikz, margin=3.141592]{standalone}
\usetikzlibrary{angles, intersections, quotes}

\begin{document}
    \begin{tikzpicture}[
myangle/.style args = {#1/#2}{angle radius=#1, angle eccentricity=#2, draw, font=\footnotesize},
myangle/.default=8mm/0.8
                        ]
\coordinate[label=$B$]      (b);
\coordinate[label=below:$C$](c) at (255:6);
\coordinate[label=below:$D$](d) at (285:6);
    \draw[name path=cbd]    (c) -- (b) -- (d) -- cycle;
    \draw[name path=A]      (b) ++ (250:6) arc (250:330:6);
    \path[name path=B]      (b) -- ++ (315:8.5);
    \draw[name path=ce,
          name intersections={of=A and B, by=e}] (c) -- (e) coordinate[label=right:E];
    \path[name path=C]      (d) -- ++ (0:4.5);
    \draw[name intersections={of=B and C, by=a}] (d) -- (a) coordinate[label=below:A];
    \draw   (b) -- (a);
% angles
\pic [myangle,"$\alpha$"] {angle = c--b--d};
\pic [myangle,"$\alpha$"] {angle = e--a--d};
    \path[name intersections={of=ce and cbd}] coordinate[label=75:F] (f) at (intersection-2);
\pic [myangle=4mm/0.5,"$\cdot$"] {angle = b--f--c};
% small circle
\draw[fill=white] (b) circle[radius=0.5mm];
    \end{tikzpicture}
\end{document}

Answer

使用纯的tikz，使用极坐标和交点：

\documentclass[11pt, tikz, margin=3.141592]{standalone}
\usetikzlibrary{angles, intersections, quotes}

\begin{document}
    \begin{tikzpicture}[
myangle/.style args = {#1/#2}{angle radius=#1, angle eccentricity=#2, draw, font=\footnotesize},
myangle/.default=8mm/0.8
                        ]
\coordinate[label=$B$]      (b);
\coordinate[label=below:$C$](c) at (255:6);
\coordinate[label=below:$D$](d) at (285:6);
    \draw[name path=cbd]    (c) -- (b) -- (d) -- cycle;
    \draw[name path=A]      (b) ++ (250:6) arc (250:330:6);
    \path[name path=B]      (b) -- ++ (315:8.5);
    \draw[name path=ce,
          name intersections={of=A and B, by=e}] (c) -- (e) coordinate[label=right:E];
    \path[name path=C]      (d) -- ++ (0:4.5);
    \draw[name intersections={of=B and C, by=a}] (d) -- (a) coordinate[label=below:A];
    \draw   (b) -- (a);
% angles
\pic [myangle,"$\alpha$"] {angle = c--b--d};
\pic [myangle,"$\alpha$"] {angle = e--a--d};
    \path[name intersections={of=ce and cbd}] coordinate[label=75:F] (f) at (intersection-2);
\pic [myangle=4mm/0.5,"$\cdot$"] {angle = b--f--c};
% small circle
\draw[fill=white] (b) circle[radius=0.5mm];
    \end{tikzpicture}
\end{document}

Question 2

只需tkz-euclide（v3.01在 CTAN 上

第一种方法：我们寻找角度CBD来做到这一点，我们称Hb为B在线CD上的投影，那么很容易看出：

Ang(Hb,B,D)+Ang(D,B,A)+Ang(Hb,A,B) = 90 or Ang(Hb,B,D)=1/2 Ang(Hb,A,B).

我们得到2.5x alpha = 90°了alpha=36°

\documentclass{standalone} 
\usepackage{tkz-euclide}
\begin{document} 
\begin{tikzpicture}

\tkzDefPoint(0,0){C} 
\tkzDefPoint(2,6){B}
% possible \tkzDefPoint[label=below:$C$](0,0){C} but don't do this
% We get D and E with a rotation
\tkzDefPointBy[rotation= center B angle 36](C) \tkzGetPoint{D} 
\tkzDefPointBy[rotation= center B angle 72](C) \tkzGetPoint{E} 
% Toget A we use an intersection of lines
\tkzInterLL(B,E)(C,D) \tkzGetPoint{A}
\tkzInterLL(C,E)(B,D) \tkzGetPoint{H}
% drawing
\tkzDrawArc[delta=10](B,C)(E)
\tkzDrawPolygon(C,B,D)
\tkzDrawSegments(D,A B,A C,E)
% angles 
\tkzMarkAngles(C,B,D E,A,D) %this is to draw the arcs
\tkzLabelAngles[pos=1.5](C,B,D E,A,D){$\alpha$}
\tkzMarkRightAngle(B,H,C)
\tkzDrawPoints(A,...,E)
% Label only now
\tkzLabelPoints[below left](C,A)
\tkzLabelPoints[below right](D)
\tkzLabelPoints[above](B,E)
\end{tikzpicture}
\end{document}

还有另一种可能性，因为三角形 CBD 是一个具有 36°、72° 和 72° 角的“黄金”三角形，并且tkz-euclide知道如何定义这种三角形。

\documentclass{standalone} 
\usepackage{tkz-euclide}
\begin{document}  
\begin{tikzpicture}
\tkzDefPoint(0,4){B}
\tkzDefPoint(8,0){A}
\tkzDefTriangle[gold](A,B) \tkzGetPoint{C}
\tkzDefTriangle[gold](B,C) \tkzGetPoint{D}
\tkzDuplicateSegment(B,C)(B,A)\tkzGetPoint{E}
\tkzInterLL(C,E)(B,D) \tkzGetPoint{F}
% drawing
\tkzDrawArc[delta=10](B,C)(E)
\tkzDrawPolygon(C,B,A)
\tkzDrawSegments(D,B C,E)
\tkzDrawPoints(A,...,F)
% % angles
\tkzFillAngles[fill=blue!10](C,B,D E,A,D)
\tkzMarkAngles(C,B,D E,A,D) %this is to draw the arcs
\tkzLabelAngles[pos=1.5](C,B,D E,A,D){$\alpha$}
\tkzMarkRightAngle[fill=blue!20](B,F,C)
% % Label only now
\tkzLabelPoints[below left](C,A)
\tkzLabelPoints[below right](D)
\tkzLabelPoints[above](B,E)
\end{tikzpicture}
\end{document}

Answer