Question

given: y || z
prove: m∠5 + m∠2 + m∠6 = 180°
diagram: line y with points l, a, m; line z with points c, b; triangle acb with angles 1,2,3 at a, 5 at c, 6 at b
table: statements (1. y || z; 2. ∠1 ≅ ∠5; 3. ∠3 ≅ ∠6; 4. m∠1 = m∠5; 5. m∠1 + m∠2 + m∠3 = m∠lam; 6. m∠1 + m∠2 + m∠3 = 180) and reasons (1. given; 2. alternate interior angles theorem; 3. alternate interior angles theorem; 4. def. of ≅; 5. angle addition postulate; 6. def. of straight angle) with drag-and-drop options for angles/lines/statements/reasons

Explanation:

Step1: Substitute \( m\angle1 \) and \( m\angle3 \)

From step 4, we know \( m\angle1 = m\angle5 \) and \( m\angle3 = m\angle6 \) (by the same logic as step 4 for \( \angle3 \) and \( \angle6 \)). Substitute \( m\angle1 \) with \( m\angle5 \) and \( m\angle3 \) with \( m\angle6 \) in the equation \( m\angle1 + m\angle2 + m\angle3 = 180^\circ \) (from step 6).
\[ m\angle5 + m\angle2 + m\angle6 = 180^\circ \]

Answer:

\( m\angle5 + m\angle2 + m\angle6 = 180^\circ \) is proven by substituting \( m\angle1 = m\angle5 \) and \( m\angle3 = m\angle6 \) into \( m\angle1 + m\angle2 + m\angle3 = 180^\circ \).