Use the image to answer the question.

Question

Two parallel horizontal lines a and b, intersecting with two parallel slanted lines e and f. A parallelogram is formed in the center of the intersecting lines.

In the diagram, a∥b
and e∥f
. Sylvia writes a proof to prove that opposite angles, ∠6
and ∠11
, are congruent in the parallelogram. Drag and drop the statements and reasons into their correct locations in the two-column proof.

(2 points)
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.
Statement Reason
1. a∥b
and e∥f
given
2. Response area Corresponding Angles Theorem
3. m∠5+m∠6=180°
and m∠9+m∠11=180°
definition of linear pair
4. m∠13+m∠6=180°
and m∠13+m∠11=180°
Response area
5. Response area Transitive Property of Equality
6. m∠6=m∠11
Response area

Answer 1

To complete the proof, we can fill in the response areas with the appropriate statements and reasons. Here's how we can arrange the proof:

| Statement | Reason | |------------------------------------------------------|------------------------------------------| | 1. a∥b and e∥f | Given | | 2. m∠5 + m∠6 = 180° and m∠9 + m∠11 = 180° | Definition of linear pair | | 3. m∠13 + m∠6 = 180° and m∠13 + m∠11 = 180° | Alternate Interior Angles Theorem | | 4. m∠6 = m∠11 | Transitive Property of Equality | | 5. m∠6 = m∠11 | Corresponding Angles Theorem |

The completed proof will look like this:

| Statement | Reason | |------------------------------------------------------|------------------------------------------| | 1. a∥b and e∥f | Given | | 2. m∠5 + m∠6 = 180° and m∠9 + m∠11 = 180° | Definition of linear pair | | 3. m∠13 + m∠6 = 180° and m∠13 + m∠11 = 180° | Alternate Interior Angles Theorem | | 4. m∠6 = m∠11 | Transitive Property of Equality | | 5. m∠6 = m∠11 | Corresponding Angles Theorem |

Make sure to place the responses in the correct areas as indicated in the proof layout above.