figures/fig-i-32.textex · 1242 bytesRaw% fig-i-32.tex — I.32: triangle angle sum (parallel through apex).
\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=1.2, line cap=round]
\coordinate (A) at (-1.5, 0);
\coordinate (B) at (2.0, 0);
\coordinate (C) at (0.5, 2.4);
\coordinate (D) at ($(C)!-1.4!(A)$); % line through C parallel to AB, on left
\coordinate (E) at ($(C)!-1.4!(B)$); % line through C parallel to AB, on right
% Triangle.
\draw[very thick] (A) -- (B) -- (C) -- cycle;
% Parallel through C, drawn long.
\draw[thin] ($(C)!-0.7!(B)$) -- ($(C)!1.5!(B)$);
% Side AB extended beyond B to F to expose exterior angle.
\coordinate (F) at ($(A)!1.4!(B)$);
\draw[thin] (B) -- (F);
% Labels.
\node[below left] at (A) {$A$};
\node[below right] at (B) {$B$};
\node[above] at (C) {$C$};
\node[below right] at (F) {$F$};
\node[above left] at ($(C)!-0.7!(B)$) {$D$};
\node[above right] at ($(C)!1.5!(B)$) {$E$};
\end{tikzpicture}
\caption{Proposition I.32. Drawing $DE$ through $C$ parallel to $AB$
makes $\angle DCA = \angle CAB$ (alternate, I.29) and $\angle ECB =
\angle ABC$ (alternate, I.29); the straight angle at $C$ then sums
the three interior angles of $\triangle ABC$ to two right angles.}
\label{fig:I.32}
\end{figure}