Question

given: (overline{nm} parallel overline{xz}) prove: (\triangle xyz sim \triangle nym) we know that side nm is (\boxed{\text{dropdown}}) to side xz. if we consider side ny the transversal for these parallel lines, we create angle pairs. using the (\boxed{\text{dropdown}}), we can state that (angle yxz) is congruent to (angle ynm). we know that angle xyz is congruent to angle (\boxed{\text{dropdown}}) by the reflexive property. therefore, triangle xyz is similar to triangle nym by the (\boxed{\text{dropdown}}) similarity theorem. (options for last dropdown: asa, sss, aa)

Explanation:

Step1: Analyze the parallel sides

Since \(\overline{NM} \parallel \overline{XZ}\), by the definition of parallel lines, they are parallel (first blank: "parallel").

Step2: Identify the angle - angle relationship

When a transversal (here \(NY\)) cuts two parallel lines (\(NM\) and \(XZ\)), the corresponding angles are congruent. This is the Corresponding Angles Postulate (second blank: "Corresponding Angles Postulate").

Step3: Use the reflexive property

The reflexive property of congruence states that an angle is congruent to itself. So \(\angle XYZ\) is congruent to \(\angle NYM\) (third blank: "\(\angle NYM\)").

Step4: Determine the similarity theorem

We have two pairs of congruent angles: \(\angle YXZ\cong\angle YNM\) and \(\angle XYZ\cong\angle NYM\). By the AA (Angle - Angle) similarity theorem, if two angles of one triangle are congruent to two angles of another triangle, the triangles are similar. So the last blank is "AA".

Answer:

1. parallel
2. Corresponding Angles Postulate
3. \(\angle NYM\)
4. AA