% fig-iii-31.tex — III.31: angle in a semicircle is right.
\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=1.1, line cap=round]
\coordinate (O) at (0, 0);
\def\r{2}
\draw[thin] (O) circle (\r);
\coordinate (A) at (-\r, 0);
\coordinate (B) at ( \r, 0);
% Diameter AB.
\draw[very thick] (A) -- (B);
\coordinate (C) at ({\r*cos(70)}, {\r*sin(70)});
\draw[very thick] (A) -- (C) -- (B);
% Right angle marker at C.
\coordinate (Cm1) at ($(C)!0.25!(A)$);
\coordinate (Cm2) at ($(C)!0.25!(B)$);
\coordinate (Cmid) at ($(Cm1)!0.5!(Cm2)$);
\draw[thin] (Cm1) -- ($(Cm1)+(Cm2)-(C)$) -- (Cm2);
\node[left] at (A) {$A$};
\node[right] at (B) {$B$};
\node[above] at (C) {$C$};
\node[below] at (O) {$O$};
\end{tikzpicture}
\caption{Proposition III.31. For any point $C$ on the circle (not at
$A$ or $B$), the inscribed angle $\angle ACB$ subtending the diameter
$AB$ is a right angle. Proof: by I.5 applied to the two isoceles
triangles $OAC$ and $OCB$, then I.32.}
\label{fig:III.31}
\end{figure}