Complete the proof.
There are three labeled lines, five labeled points, four labeled angles, and two line segments.
Line ℓ and line n appear horizontal, with line ℓ above line n.
Line t starts below both ℓ and n and travels up and to the right intersecting n at point C and then intersecting ℓ at point B.
Point A is on line ℓ and to the right of line t.
Point D is on line n and to the right of line t.
Point E is near the center of the space that is below line ℓ, above line n and to the right of line t.
A line segment connects point B to point E.
A line segment connects point C to point E. This creates the triangle BCE.
∠ABE is labeled 1.
∠CBE is labeled 2.
∠BCE is labeled 3.
∠ECD is labeled 4.
Given:
m∠2 + m∠3 = 90°
BE bisects ∠ABC.
CE bisects ∠BCD.
Prove: ℓ ∥ n
Statements Reasons
1. m∠2 + m∠3 = 90°
1. Given
2. BE bisects ∠ABC; CE bisects ∠BCD.
2. Given
3. m∠1 = m∠2; m∠3 = m∠4
3. If a ray bisects an ∠, it forms two ∠s of equal measure.
4. m∠1 + m∠4 = 90°
4. Substitution
5. m∠1 + m∠2 + m∠3 + m∠4 = 180°
5. Addition Property of Equality
6. m∠ABC = m∠1 + m∠2; m∠BCD = m∠3 + m∠4
6. Angle-Addition Postulate
7. m∠ABC + m∠BCD = 180°
7. Substitution
8. ∠ABC is supplementary to ∠BCD.
8. If the sum of measures of two ∠s is 180°, the ∠s are supplementary.
9. ℓ ∥ n
9. If two lines are cut by a transversal so that interior ∠s on the same side of the transversal are supplementary, then the lines are ∥.