我正在考虑复制这种几何演示风格；
答案1
Slyusarev Sergey 使用 Metapost 在 ConTeXt 中完整复制了 Byrne 的 Euclid（您也可以在 LaTeX 中使用 Metapost），可在此处找到https://github.com/jemmybutton/byrne-euclid。您可以从发布部分。
最重要的文件是https://github.com/jemmybutton/byrne-euclid/blob/master/byrne.mp其中包含所有图片的 metapost 定义（即主文件中使用的宏）。最基本绘图的宏以 为前缀by，例如byNamedLine。
\startProposition[title={Prop XXXIV. Theor.}, reference=prop:I.XXXIV]
\defineNewPicture{
pair A, B, C, D, d[];
d1 := (5/2u, 0);
d2 := (-7/8u, -3u);
A := (0, 0);
B := A shifted d1;
C := A shifted d2;
D := C shifted d1;
byAngleDefine(B, A, D, byblue, 0);
byAngleDefine(D, A, C, byred, 0);
byAngleDefine(C, D, A, byyellow, 0);
byAngleDefine(A, D, B, byred, 0);
byAngleDefine(A, C, D, byblack, 0);
byAngleDefine(D, B, A, byblack, 0);
draw byNamedAngleResized();
draw byLine(A, D, byblack, 0, 0);
byLineDefine(A, B, byred, 0, 0);
byLineDefine(C, D, byred, 1, 0);
byLineDefine(A, C, byyellow, 0, 0);
byLineDefine(B, D, byblue, 0, 0);
draw byNamedLineSeq(0)(AB,BD,CD,AC);
draw byLabelsOnPolygon(A, B, D, C)(0, 0);
}
\drawCurrentPictureInMargin
\problemNP{T}{he}{opposite sides and angles of any parallelogram are equal, and the diagonal (\drawUnitLine{AD}) divides it into two equal parts.}
\startCenterAlign
Since $\left\{\eqalign{\drawAngle{BAD} &= \drawAngle{CDA}\cr\drawAngle{DAC} &= \drawAngle{ADB}\cr}\right\}$ \inprop[prop:I.XXIX]\\
and \drawUnitLine{AD} common to the two triangles.
$\therefore \left\{\eqalign{\drawUnitLine{AB} &= \drawUnitLine{CD}\cr \drawUnitLine{AC} &= \drawUnitLine{BD}\cr \drawAngle{B} &= \drawAngle{C}\cr}\right\}$ \inprop[prop:I.XXVI]\\
and $\drawAngle{BAD,DAC} = \drawAngle{CDA,ADB}$ \inax[ax:I.II].
\stopCenterAlign
Therefore the opposite sides and angles of the parallelogram are equal: and as the triangles \drawLine{AD,CD,AC} and \drawLine{AB,BD,AD} are equal in every respect \inprop[prop:I.IV], the diagonal divides the parallelogram into two equal parts.
\qed
\stopProposition
这是你在问题中展示的命题的代码。它呈现为
答案2
除了包含图形之外，正如 Ulrike 所提到的，您还可以只放tikzpictures。（这主要是出于我自己的好奇心而写的。我知道字体并不完全匹配。）这样您就可以在文本中的小图片中获得平行四边形的精确角度。甚至，Ti钾Z 为您计算角度，即如果您选择不同的平行四边形，角度将自动调整。
\documentclass{article}
\usepackage[right = 3in]{geometry}
\usepackage[utf8]{inputenc}
\usepackage{chancery}
\usepackage[T1]{fontenc}
\usepackage{lettrine}
\usepackage{GoudyIn} % use texdoc cfr-initials to see alternatives
\usepackage{amssymb}
\usepackage{tikzpagenodes}
\usetikzlibrary{calc}
\renewcommand{\LettrineFontHook}{\GoudyInfamily{}}
\LettrineTextFont{\itshape}
\setcounter{DefaultLines}{3}
\setlength{\DefaultNindent}{0pt}
\setlength{\DefaultFindent}{1pt}
\newcommand\CenterObject[1]{\vcenter{\hbox{#1}}}
\newcommand\CenterTikz[1]{\vcenter{\hbox{\begin{tikzpicture}
#1
\end{tikzpicture}}}}
\definecolor{dull}{RGB}{249,243,217}
\pagecolor{dull}
\begin{document}
\begin{tikzpicture}[overlay,remember picture]
\path ([xshift=-1.5cm,yshift=-1cm]current page.east|-current page text area.north)
coordinate (tr) ++ (180:3cm) coordinate (tl)
++ (-110:3cm) coordinate (bl) ++ (0:3cm) coordinate (br);
\path[fill=blue] let \p1=($(br)-(tl)$), \n1={atan2(\y1,\x1)} in
\pgfextra{\pgfmathsetmacro{\myAngleTLA}{\n1}\xdef\myAngleTLA{\myAngleTLA}}
(tl) -- ++ (1,0) arc[start angle=0,end angle=\myAngleTLA,radius=1];
\path[fill=red] let \p1=($(bl)-(tl)$), \n1={atan2(\y1,\x1)} in
\pgfextra{\pgfmathsetmacro{\myAngleTLB}{\n1}\xdef\myAngleTLB{\myAngleTLB}}
(tl) -- ++ (\myAngleTLA:1) arc[start angle=\myAngleTLA,end angle=\myAngleTLB,radius=1];
\path[fill=yellow!50!orange]
(br) -- ++ (-1,0) arc[start angle=180,end angle=180+\myAngleTLA,radius=1];
\path[fill=red]
(br) -- ++ (180+\myAngleTLA:1) arc[start angle=180+\myAngleTLA,end angle=180+\myAngleTLB,radius=1];
\path[fill=black] (tr) -- ++ (180:1)
arc[start angle=180,end angle=360+\myAngleTLB,radius=1]
(bl) -- ++ (0:1)
arc[start angle=0,end angle=\myAngleTLB+180,radius=1];
\path[ultra thick,line cap=round] (tr) edge[red] (tl)
(tl) edge[yellow!50!orange] (bl)
(bl) edge[red,densely dashed] (br)
(tr) edge[blue] (br) (tl) edge (br);
\end{tikzpicture}
\lettrine{T}{HE} opposite sides and angles of any parallelogram are equal, and
the diagonal $(\CenterTikz{\draw[very thick] (0,0) --(2em,0);})$ divides it into equal parts.\bigskip
\noindent
Since $\left\{\begin{array}{@{}r@{\,}c@{\,}l@{}}
\CenterTikz{%
\fill[blue] (0,0) -- (0.5,0) arc[start angle=0,end angle=\myAngleTLA,radius=0.5];
} &= &
\CenterTikz{%
\fill[yellow!50!orange] (0,0) -- (-0.5,0)
arc[start angle=180,end angle=180+\myAngleTLA,radius=0.5];
}\\[2ex]
\CenterTikz{%
\fill[red] (0,0) -- (\myAngleTLA:0.5) arc[start angle=\myAngleTLA,end angle=\myAngleTLB,radius=0.5];
} &= &
\CenterTikz{%
\fill[red] (0,0) -- (180+\myAngleTLA:0.5) arc[start angle=180+\myAngleTLA,end angle=180+\myAngleTLB,radius=0.5];
}\\
\end{array}\right\}$ and $\vcenter{\hbox{\begin{tikzpicture}\draw[very thick] (0,0) --
(2em,0);\end{tikzpicture}}}$ are common to the two triangles,
\[
\therefore\left\{\begin{array}{@{}r@{\,}c@{\,}l@{}}
\CenterTikz{\draw[very thick,red] (0,0) --(2em,0);}
&=&
\CenterTikz{\draw[very thick,red,densely dashed] (0,0) --(2em,0);}\\
\CenterTikz{\draw[very thick,yellow!50!orange] (0,0) --(2em,0);}
&=&
\CenterTikz{\draw[very thick,blue] (0,0) --(2em,0);}\\
\CenterTikz{\path[fill=black] (0,0) -- ++ (180:0.5)
arc[start angle=180,end angle=360+\myAngleTLB,radius=0.5];}
&=&
\CenterTikz{\path[fill=black] (0,0) -- ++ (0:0.5)
arc[start angle=0,end angle=180+\myAngleTLB,radius=0.5];}\\
\end{array}\right\}
\]
and
\[
\CenterTikz{%
\fill[blue] (0,0) -- (0.5,0) arc[start angle=0,end angle=\myAngleTLA,radius=0.5];
\fill[red] (0,0) -- (\myAngleTLA:0.5) arc[start angle=\myAngleTLA,end angle=\myAngleTLB,radius=0.5];
}
=
\CenterTikz{%
\fill[yellow!50!orange] (0,0) -- (-0.5,0)
arc[start angle=180,end angle=180+\myAngleTLA,radius=0.5];
\fill[red] (0,0) -- (180+\myAngleTLA:0.5) arc[start angle=180+\myAngleTLA,end angle=180+\myAngleTLB,radius=0.5];
}\;.
\]
\end{document}
