Points [tex]\( A, B \)[/tex], and [tex]\( C \)[/tex] form a triangle. 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]\(\overline{DE}\)[/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] & \\
\hline
[tex]\(m\angle 1 = m\angle 4\)[/tex] and [tex]\(m\angle 3 = m\angle 5\)[/tex] & [tex]\(\square\)[/tex] \\
\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}



Answer :

Certainly! Let's complete the proof to show that the sum of the interior angles of [tex]\(\triangle ABC\)[/tex] is [tex]\(180^{\circ}\)[/tex].

[tex]\[ \begin{tabular}{|l|l|} \hline Statement & Reason \\ \hline Points $A, B$, and $C$ form a triangle. & Given \\ \hline Let $\overline{DE}$ be a line passing through $B$ and parallel to $\overline{AC}$ & Definition of parallel lines \\ \hline $\angle 3 \cong \angle 5$ and $\angle 1 \cong \angle 4$ & Alternate interior angles are congruent \\ \hline $m\angle 1 = m\angle 4$ and $m\angle 3 = m\angle 5$ & Congruent angles have equal measures \\ \hline $m\angle 4 + m\angle 2 + m\angle 5 = 180^{\circ}$ & Angle addition and definition of a straight line \\ \hline $m\angle 1 + m\angle 2 + m\angle 3 = 180^{\circ}$ & Substitution \\ \hline \end{tabular} \][/tex]

Here’s a detailed explanation of each step:

1. Points [tex]\(A, B, and C\)[/tex] form a triangle.
- This is given.

2. Let [tex]\(\overline{DE}\)[/tex] be a line passing through [tex]\(B\)[/tex] and parallel to [tex]\(\overline{AC}\)[/tex].
- By the definition of parallel lines, we draw a line [tex]\(\overline{DE}\)[/tex] that is parallel to [tex]\(\overline{AC}\)[/tex] and passes through point [tex]\(B\)[/tex].

3. [tex]\(\angle 3 \cong \angle 5\)[/tex] and [tex]\(\angle 1 \cong \angle 4\)[/tex].
- This is because [tex]\(\angle 3\)[/tex] and [tex]\(\angle 5\)[/tex] are alternate interior angles formed by the transversal [tex]\(\overline{BC}\)[/tex] through the parallel lines [tex]\(\overline{DE}\)[/tex] and [tex]\(\overline{AC}\)[/tex]. Similarly, [tex]\(\angle 1\)[/tex] and [tex]\(\angle 4\)[/tex] are alternate interior angles formed by the transversal [tex]\(\overline{AB}\)[/tex] through the parallel lines [tex]\(\overline{DE}\)[/tex] and [tex]\(\overline{AC}\)[/tex]. Alternate interior angles between parallel lines are congruent.

4. [tex]\(m\angle 1 = m\angle 4\)[/tex] and [tex]\(m\angle 3 = m\angle 5\)[/tex].
- Since [tex]\(\angle 1 \cong \angle 4\)[/tex] and [tex]\(\angle 3 \cong \angle 5\)[/tex], their measures are equal.

5. [tex]\(m\angle 4 + m\angle 2 + m\angle 5 = 180^{\circ}\)[/tex].
- As [tex]\(\overline{DE}\)[/tex] forms a straight line, the angles along this straight line add up to [tex]\(180^{\circ}\)[/tex]. Therefore, [tex]\(m\angle 4 + m\angle 2 + m\angle 5 = 180^{\circ}\)[/tex].

6. [tex]\(m\angle 1 + m\angle 2 + m\angle 3 = 180^{\circ}\)[/tex].
- By substituting [tex]\(m\angle 4\)[/tex] with [tex]\(m\angle 1\)[/tex] and [tex]\(m\angle 5\)[/tex] with [tex]\(m\angle 3\)[/tex] from previous equality statements, we get [tex]\(m\angle 1 + m\angle 2 + m\angle 3 = 180^{\circ}\)[/tex].

Thus, the sum of the interior angles of [tex]\(\triangle ABC\)[/tex] is [tex]\(180^{\circ}\)[/tex].