Drag and drop an answer to each box to correctly complete the proof. Given: m∥n , m∠1=50∘ , m∠2=48∘ , and line s bisects ∠ABC . Prove: m∠3=49∘ Line m parallel to line n. Line t passing through both lines. There are four angles formed by lines m and t intersecting at point E. The upper left angle is angle D E F with point D on line m and point F on line t. Angle D E F is separated by a ray forming angles 1 and 2, where angle 1 is the left angle of those two angles. There are four angles formed by lines n and t intersecting at point B. The lower right angle is angle A B C with point A on line t and point C on line n. Line s passes through point B separating angle A B C into two angles labeled 4 and 5 with angle 4 being on the left side. Line s also separates the upper left angle into two angles, and the angle on the right side is labeled 3. Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse. It is given that m∥n , m∠1=50∘ , m∠2=48∘ , and line s bisects ∠ABC . By the Response area, m∠DEF=98∘ . Because Response area angles formed by two parallel lines and a transversal are congruent, ∠DEF≅∠ABC , so m∠ABC=98∘ . By the Response area, angles 4 and 5 are congruent, and m∠4 is half m∠ABC . So m∠4=49∘ . Because vertical angles are congruent, ∠3≅∠4 . Finally, m∠3=m∠4 by the angle congruence postulate, so m∠3=49∘ by the Response area.