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Given: a ∥ b and ∠1 ≅ ∠3
Prove: e ∥ f
Horizontal and parallel lines e and f are intersected by parallel lines a and b. At the intersection of lines a and e, the bottom left angle is angle 1. At the intersection of lines b and e, the uppercase right angle is angle 2. At the intersection of lines f and b, the bottom left angle is angle 3 and the bottom right angle is angle 4.
We know that angle 1 is congruent to angle 3 and that line a is parallel to line b because they are given. We see that __________ by the alternate exterior angles theorem. Therefore, angle 2 is congruent to angle 3 by the transitive property. So, we can conclude that lines e and f are parallel by the converse alternate exterior angles theorem.
Which information is missing in the paragraph proof?
∠2 ≅ ∠4
∠1 ≅ ∠2
∠2 ≅ ∠3
∠1 ≅ ∠4
Prove: e ∥ f
Horizontal and parallel lines e and f are intersected by parallel lines a and b. At the intersection of lines a and e, the bottom left angle is angle 1. At the intersection of lines b and e, the uppercase right angle is angle 2. At the intersection of lines f and b, the bottom left angle is angle 3 and the bottom right angle is angle 4.
We know that angle 1 is congruent to angle 3 and that line a is parallel to line b because they are given. We see that __________ by the alternate exterior angles theorem. Therefore, angle 2 is congruent to angle 3 by the transitive property. So, we can conclude that lines e and f are parallel by the converse alternate exterior angles theorem.
Which information is missing in the paragraph proof?
∠2 ≅ ∠4
∠1 ≅ ∠2
∠2 ≅ ∠3
∠1 ≅ ∠4
Answers
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Answered by
GPT-5 mini
∠1 ≅ ∠2
Reason: a ∥ b and e is a transversal, so ∠1 and ∠2 are alternate exterior angles and hence congruent. Combining ∠1 ≅ ∠2 with the given ∠1 ≅ ∠3 gives ∠2 ≅ ∠3, and by the converse alternate exterior angles theorem e ∥ f.
Reason: a ∥ b and e is a transversal, so ∠1 and ∠2 are alternate exterior angles and hence congruent. Combining ∠1 ≅ ∠2 with the given ∠1 ≅ ∠3 gives ∠2 ≅ ∠3, and by the converse alternate exterior angles theorem e ∥ f.
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