Here’s the corrected version of the solution to prove that the sum of the interior angles of [tex]\(\triangle ABC\)[/tex] is [tex]\(180^\circ\)[/tex]:
\begin{tabular}{|c|c|}
\hline
Statement & Reason \\
\hline
Points [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] form a triangle. & Given \\
\hline
Let [tex]\(\overline{DE}\)[/tex] be a line passing through [tex]\(B\)[/tex] and parallel to [tex]\(\overline{AC}\)[/tex]. & Definition of parallel lines \\
\hline
[tex]\(\angle 3 \cong \angle 5\)[/tex] and [tex]\(\angle 1 \cong \angle 4\)[/tex] & Alternate interior angles are congruent \\
\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
\end{tabular}
By these steps, you can see we are using geometric properties and definitions to show that the interior angles of [tex]\(\triangle ABC\)[/tex] add up to [tex]\(180^\circ\)[/tex].
