Sure, let's complete the statements to prove that the sum of the interior angles of [tex]\(\triangle ABC\)[/tex] is [tex]\(180^\circ\)[/tex].
\begin{tabular}{|l|l|}
\hline Statement & Reason \\
\hline Points [tex]$A, B$[/tex], and [tex]$C$[/tex] form a triangle. & given \\
\hline Let [tex]$\overline{D E}$[/tex] be a line passing through [tex]$B$[/tex] and parallel to [tex]$\overline{A C}$[/tex] & definition of parallel lines \\
\hline[tex]$\angle 3 \cong \angle 5$[/tex] and [tex]$\angle 1 \cong \angle 4$[/tex] & alternate interior angles or corresponding angles are equal when lines are parallel \\
\hline [tex]$m \angle 1= m \angle 4$[/tex] and [tex]$m \angle 3= m \angle 5$[/tex] & definition of congruent angles \\
\hline [tex]$m \angle 4+ m \angle 2+ m \angle 5=180^{\circ}$[/tex] & angle addition and definition of a straight line \\
\hline [tex]$m \angle 1+ m \angle 2+ m \angle 3=180^{\circ}$[/tex] & substitution \\
\hline
\end{tabular}
So, by following these logical steps, we can confirm that the sum of the interior angles of [tex]\(\triangle ABC\)[/tex] is indeed [tex]\(180^\circ\)[/tex].
