Let's go through a step-by-step solution to prove that the sum of the interior angles of [tex]\( \triangle ABC \)[/tex] is [tex]\( 180^\circ \)[/tex].
1. Points [tex]\( A, B \)[/tex], and [tex]\( C \)[/tex] form a triangle
Reason: Given
2. Let [tex]\( \overline{DE} \)[/tex] be a line passing through [tex]\( B \)[/tex] and parallel to [tex]\( \overline{AC} \)[/tex]
Reason: Definition of parallel lines
3. [tex]\( \angle 3 \cong \angle 5 \)[/tex] and [tex]\( \angle 1 \cong \angle 4 \)[/tex]
Reason: Alternate Interior Angles Theorem (When a transversal intersects two parallel lines, alternate interior angles are congruent)
4. [tex]\( m\angle 1 = m\angle 4 \)[/tex] and [tex]\( m\angle 3 = m\angle 5 \)[/tex]
Reason: Congruent angles have equal measures
5. [tex]\( m\angle 4 + m\angle 2 + m\angle 5 = 180^\circ \)[/tex]
Reason: Angle addition and definition of a straight line (The sum of the angles on a straight line is [tex]\( 180^\circ \)[/tex])
6. [tex]\( m\angle 1 + m\angle 2 + m\angle 3 = 180^\circ \)[/tex]
Reason: Substitution (Since [tex]\( m\angle 1 = m\angle 4 \)[/tex] and [tex]\( m\angle 3 = m\angle 5 \)[/tex], substitute these into the previous equation)
Thus, we have proven that the sum of the interior angles of [tex]\( \triangle ABC \)[/tex] is [tex]\( 180^\circ \)[/tex].
