Fred and victoria provide the following proofs for vertical angles to be equal:
a line pq is shown cut by a transversal t. 1, 2, 3, 4 are marked clockwise as the four angles formed by the transversal on the segment pq
fred's proof: angle 2 + angle 3 = 180° (t is a straight line)
angle 1 + angle 2 = 180° (pq is a straight line)
therefore, angle 1 + angle 2 = angle 2 + angle 3 (transitive property of equality)
hence, angle 1 = angle 3 (subtraction property of equality)
victoria's proof: angle 1 + angle 4 = 180° (t is a straight line)
angle 1 + angle 2 =180° (pq is a straight line)
therefore, angle 1 + angle 2 = angle 1 + angle 4 (transitive property of equality)
hence, angle 2 = angle 4 (subtraction property of equality)
which statement is correct?
both fred's and victoria's proofs are correct.
both fred's and victoria's proofs are incorrect.
only fred's proof is correct.
only victoria's proof is correct.

the answer is A. both fred's and victoria's proofs are correct. I just took the exam and got it right hope this helps!

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

to have more of what you want like money

step-by-step explanation:

save it by not spending on unnecessary stuff and have more