Which point is not a solution to the system of inequalities? Thank you:)))
1 answer:
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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.