What are the missing parts that correctly complete the proof? given: point q is on the perpendicular bisector of segment m n. prove: point q is equidistant from the endpoints of segment m n. art: a horizontal line segment m n with x as the midpoint is drawn. a vertical line x q is drawn. q is above the horizontal line. the angle q x n is labeled as a right angle. the line segments m x and n x are labeled with a single tick mark. a dotted line is used to connect point q with point m. another dotted line is used to connect point q with point n. drag the answers into the boxes to correctly complete the proof. statement reason 1. point q is on the perpendicular bisector of mn¯¯¯¯¯¯¯. given 2. definition of bisector 3. ∠qxm and ∠qxn are right angles. definition of perpendicular 4. ∠qxm≅∠qxn 5. reflexive property of congruence 6. △qxm≅△qxn 7. corresponding parts of congruent triangles are congruent. 8. point q is equidistant from the endpoints of mn¯¯¯¯¯¯¯. definition of equidistant definition of bisectorgivenasa congruence postulatesas congruence postulateqm¯¯¯¯¯¯≅qn¯¯¯¯¯¯mx¯¯¯¯¯¯≅ nx¯¯¯¯¯¯qx¯¯¯¯¯¯≅qx¯¯¯¯¯¯all right angles are congruent.

Point Q is equidistant from the endpoints of segment M N by  Definition of equidistant.

Step-by-step explanation:

1. Point Q is on the perpendicular bisector of MN¯¯¯¯¯¯¯. Given

2.MX¯¯¯¯¯¯≅NX Definition of bisector

3. ∠QXM and ∠QXN are right angles. Definition of perpendicular

4. ∠QXM≅∠QXNAll right angles are congruent.

5.QX¯¯¯¯¯¯≅QX Reflexive Property of Congruence

6. △QXM≅△QXN Congruence PostulateSAS

7.QM¯¯¯¯¯¯≅QNCorresponding parts of congruent triangles are congruent.

8.Point Q is equidistant from the endpoints of MN¯¯¯¯¯¯¯. Definition of equidistant

answer: a clock

step-by-step explanation:

∠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.