Two lines are intersected by a third line. Two horizontal parallel lines are intersected by a third line. The third line intersects the top line and forms 4 angles. Labeled clockwise, from uppercase left, the angles are 1, 2, 4, 3. The third line intersects the bottom line and forms 4 angles. Labeled clockwise, from uppercase left, the angles are 5, 6, 8, 7. If Angle2 Is-congruent-to Angle6, which must be true about Angle2? Angle2 Is-congruent-to Angle5 Angle2 is complementary to Angle5. mAngle2 = mAngle8 Angle2 is supplementary to Angle8.

∠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.

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point