Which of the following would be a line of reflection that would map ABCD onto itself? Square ABCD with C at 1 comma 3, D at 1 comma 1, A at 3 comma 1, and B at 3 comma 3. y = 1 2x + y = 2 x + y = 4 x + y = 1

  x + y = 4

Step-by-step explanation:

Any line of symmetry will be a suitable line of reflection. Lines of symmetry connect the midpoints of opposite sides, and connect opposite vertices.

The line AC will be a line of reflection that maps the square to itself. Its equation is ...

  x + y = 4 . . . . . . . 3rd choice

∠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.

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