figures/fig-i-5.textex · 1139 bytesRaw% fig-i-5.tex — I.5: pons asinorum (base angles of an isoceles triangle).
\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=1.0, line cap=round]
\coordinate (A) at (0, 3);
\coordinate (B) at (-1.5, 0);
\coordinate (C) at (1.5, 0);
% Equal sides AB and AC, extended to F and G.
\coordinate (F) at ($(A)!2.0!(B)$);
\coordinate (G) at ($(A)!2.0!(C)$);
\draw[thin] (A) -- (F);
\draw[thin] (A) -- (G);
\draw[very thick] (B) -- (C);
% Points D on BF, E on CG with BD = CE.
\coordinate (D) at ($(B)!0.5!(F)$);
\coordinate (E) at ($(C)!0.5!(G)$);
\draw[very thick, dashed] (B) -- (E);
\draw[very thick, dashed] (C) -- (D);
% Labels.
\node[above] at (A) {$A$};
\node[left] at (B) {$B$};
\node[right] at (C) {$C$};
\node[below left] at (D) {$D$};
\node[below right] at (E) {$E$};
\node[below left] at (F) {$F$};
\node[below right] at (G) {$G$};
\end{tikzpicture}
\caption{Proposition I.5. With $AB = AC$ and $BD = CE$ on the extensions,
triangles $ABE$ and $ACD$ are congruent (SAS, by I.4), whence the base
angles $\angle ABC = \angle ACB$.}
\label{fig:I.5}
\end{figure}