如果已知角的顶点和大小，则定义构成角的角底的两个点

Question

这幅图究竟该如何画呢？

首先我定义了点A，那么我该如何定义点B和C才能形成正确的图像呢？

\documentclass[tikz,border=5]{standalone}
\usepackage{amsmath,amssymb,fouriernc,CrimsonPro}
\usetikzlibrary{calc,intersections,angles}
\begin{document}
\begin{tikzpicture}[font=\footnotesize,line join=round]
\coordinate (O) at (0:0) coordinate (X) at (0:2.5)
coordinate (Y) at (90:2.5) coordinate (A) at (46:2);
\coordinate (P) at ($(O)!(A)!(Y)$);
\coordinate[label=left:$B$] (B) at ($(A)!6.8cm!(P)$);
\coordinate (Q) at ([shift={(30:2)}]234:6.3);
\coordinate (S) at ([shift={(B)}]-35:7);
\path[name path=L] (O) circle (2);
\path[name path=g1] (A)--(B);
\path[name path=g2] (A)--(Q);
\path[name path=g3] (S)--(B);
\path [name intersections={of = L and g1, by={A,F}}];
\path [name intersections={of = L and g3, by={D,E}}];
\coordinate (T) at ($(F)!5.5cm!(D)$);
\path[name path=g4] (A)--(Q);
\path[name path=g5] (F)node[above,xshift=-.5mm]{$F$}--(T);
\path [name intersections={of = g4 and g5, by={C}}];
\pic[draw=red!60!black!40!orange,angle radius=8mm,pic text=$54^\circ$,angle eccentricity=.7] {angle={B--A--C}};
\pic[draw=red!60!black!40!orange,angle radius=9mm,pic text=$35^\circ$,pic text options={shift={(0,.03)}},angle eccentricity=.7] {angle={E--B--A}};
\pic[draw=red!60!black!40!orange,angle radius=9mm,pic text=$y$,pic text options={shift={(-.05,0)}},angle eccentricity=.75] {angle={A--C--F}};
\pic[draw=red!60!black!40!orange,angle radius=4mm,pic text=$x$,pic text options={shift={(-.05,0)}},angle eccentricity=.55] {angle={C--F--A}};
\draw[thick] (O) circle (2) (E)node[below,xshift=.5mm]{$E$}--(B)--(A)node[right,shift={(-.15,.1)}]{$A$}--(C) node[below]{$C$}--(F);
\node[left,shift={(.05,-.1)}] at (D) {$D$};
\end{tikzpicture}
\end{document}

Answer 1

你可以

1. 确定该点为从( )角E开始的射线与圆的交点，A180+alphaalpha=54
2. 找到从左边B发出的射线与从下方发出的射线的交点，并且角度为（），AE180-betabeta=35
3. 得到与圆D的交点，以及与圆的交点，E--BFA--B
4. 最后根据和C的交点确定。A--ED--F

结果如下。

\documentclass[tikz,border=5]{standalone}
\usepackage{amsmath,amssymb,fouriernc,CrimsonPro}
\usetikzlibrary{calc,intersections,angles}
\begin{document}
\begin{tikzpicture}[font=\footnotesize,line join=round,
    declare function={R=2;alpha=54;beta=35;}]
\coordinate (O) at (0:0) coordinate (X) at (0:2.5)
coordinate (Y) at (90:2.5) coordinate[label=above right:$A$] (A) at (46:R);
\path[overlay,name path=aux1] (A) +(180+54:1) -- ++ (180+54:5);
\draw[name path=L] (O) circle[radius=R];
\path[name intersections={of=aux1 and L,by=E}]; 
\path  (A) ++ (180:3) coordinate (aux1) (E) ++ (180-beta:3) coordinate (aux2)
(intersection of A--aux1 and E--aux2) coordinate[label=left:$B$] (B);
\path[name path=aux2] (E) -- (B);
\path[name intersections={of=aux2 and L,by=D}];
\path[name path=aux3] (A)  ++ (-1,0)-- (B);
\path[name intersections={of=aux3 and L,by=F}];
\path  (D) node[below left]{$D$} (E) node[below]{$E$}
    (F) node[above left]{$F$} 
    (intersection of A--E and F--D) coordinate[label=below:$C$](C);
\draw  (A) -- (C) -- (F) (A) -- (B) -- (E) -- cycle;
\pic[draw=red!60!black!40!orange,angle radius=9mm,pic text=$54^\circ$,angle eccentricity=.6] {angle={B--A--C}};
\pic[draw=red!60!black!40!orange,angle radius=10mm,pic text=$35^\circ$,pic text options={shift={(0,.03)}},angle eccentricity=.7] {angle={E--B--A}};
\pic[draw=red!60!black!40!orange,angle radius=6mm,pic text=$y$,angle eccentricity=.65] {angle={A--C--F}};
\pic[draw=red!60!black!40!orange,angle radius=4mm,pic text=$x$,angle eccentricity=.55] {angle={C--F--A}};
\end{tikzpicture}
\end{document}