because 3x6=18 so it will be the same problem
∠1≅∠2 by the alternate exterior angles theorem.
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.
a solution is a point of intersection and graph b neither of the lines intersect.