We know that side NM is parallel to side XZ. If we consider side NY the transversal for these parallel lines, we create angle pa
irs. Using the corresponding angles theorem, we can state that ∠YXZ is congruent to ∠YNM. We know that angle XYZ is congruent to angle what by the reflexive property. Therefore, triangle XYZ is similar to triangle NYM by the what similarity theorem.
2 answers:
See the picture attached.
We know:
NM // XZ
NY = transversal line
∠YXZ ≡ ∠YNM
1) <span>We know that ∠XYZ is congruent to ∠NYM by the reflexive property.</span>
The reflexive property states that any shape is congruent to itself.
∠NYM is just a different way to call ∠XYZ using different vertexes, but the sides composing the two angles are the same.
Hence, ∠XYZ ≡ <span>∠NYM</span> by the reflexive property.
2) Δ<span>XYZ is similar to ΔNYM by the AA (angle-angle) similarity theorem
The AA similarity theorem states that if two triangles have a pair of corresponding angles congruent, then the two triangles are similar.
Consider </span>Δ<span>XYZ and ΔNYM:
</span>∠YXZ ≡ <span>∠YNM
</span>∠XYZ ≡ ∠NYM
Hence, ΔXYZ is similar to ΔNYM by the AA similarity theorem.
We know:
NM // XZ
NY = transversal line
∠YXZ ≡ ∠YNM
1) <span>We know that ∠XYZ is congruent to ∠NYM by the reflexive property.</span>
The reflexive property states that any shape is congruent to itself.
∠NYM is just a different way to call ∠XYZ using different vertexes, but the sides composing the two angles are the same.
Hence, ∠XYZ ≡ <span>∠NYM</span> by the reflexive property.
2) Δ<span>XYZ is similar to ΔNYM by the AA (angle-angle) similarity theorem
The AA similarity theorem states that if two triangles have a pair of corresponding angles congruent, then the two triangles are similar.
Consider </span>Δ<span>XYZ and ΔNYM:
</span>∠YXZ ≡ <span>∠YNM
</span>∠XYZ ≡ ∠NYM
Hence, ΔXYZ is similar to ΔNYM by the AA similarity theorem.
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