Drag and drop an answer to each box to correctly complete the proof.
Given: m∥nm∥n , m∠1=50∘m∠1=50∘ , m∠2=48∘m∠2=48∘ , and line s bisects ∠ABC∠ABC .
It is given that m∥nm∥n , m∠1=50∘m∠1=50∘ , m∠2=48∘m∠2=48∘ , and line s bisects ∠ABC∠ABC . By the , m∠DEF=98∘m∠DEF=98∘ . Because angles formed by two parallel lines and a transversal are congruent, ∠DEF≅∠ABC∠DEF≅∠ABC , so m∠ABC=98∘m∠ABC=98∘ . By the , angles 4 and 5 are congruent, and m∠4m∠4 is half m∠ABCm∠ABC . So the measure of m∠4=49∘m∠4=49∘ . Because vertical angles are congruent, ∠3≅∠4∠3≅∠4 . Finally, m∠3=m∠4m∠3=m∠4 by the angle congruence postulate, so m∠3=49∘m∠3=49∘ by the .
Here is a proof
Angle DEF = angle ABC
A transversal cutting two parallel lines makes alternate exterior angles equal
Sum of measures of angles 1 and 2 = 50° + 48° = 98° = measure of angle DEF = measure of angle ABC
But since s bisects angle ABC, then angle 4 = 98 / 2 = 49°
But angles 3 and 4 are vertical angles....therefore.... measure of angle 3 = 49°