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Figure
1% fig-i-5.tex — I.5: pons asinorum (base angles of an isoceles triangle).
2\begin{figure}[H]
3\centering
4\begin{tikzpicture}[scale=1.0, line cap=round]
5 \coordinate (A) at (0, 3);
6 \coordinate (B) at (-1.5, 0);
7 \coordinate (C) at (1.5, 0);
8 % Equal sides AB and AC, extended to F and G.
9 \coordinate (F) at ($(A)!2.0!(B)$);
10 \coordinate (G) at ($(A)!2.0!(C)$);
11 \draw[thin] (A) -- (F);
12 \draw[thin] (A) -- (G);
13 \draw[very thick] (B) -- (C);
14 % Points D on BF, E on CG with BD = CE.
15 \coordinate (D) at ($(B)!0.5!(F)$);
16 \coordinate (E) at ($(C)!0.5!(G)$);
17 \draw[very thick, dashed] (B) -- (E);
18 \draw[very thick, dashed] (C) -- (D);
19 % Labels.
20 \node[above] at (A) {$A$};
21 \node[left] at (B) {$B$};
22 \node[right] at (C) {$C$};
23 \node[below left] at (D) {$D$};
24 \node[below right] at (E) {$E$};
25 \node[below left] at (F) {$F$};
26 \node[below right] at (G) {$G$};
27\end{tikzpicture}
28\caption{Proposition I.5. With $AB = AC$ and $BD = CE$ on the extensions,
29triangles $ABE$ and $ACD$ are congruent (SAS, by I.4), whence the base
30angles $\angle ABC = \angle ACB$.}
31\label{fig:I.5}
32\end{figure}
33