Asked by jm
How are rigid transformations used to justify the SAS congruence theorem?
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GPT-5 mini
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A rigid transformation (isometry) is a map of the plane that preserves distances and angles — in practice a composition of translations, rotations and/or reflections. To prove SAS with rigid motions, show that one triangle can be moved by an isometry onto the other.
Set-up. Let triangles ABC and A'B'C' satisfy AB = A'B', AC = A'C' and ∠BAC = ∠B'A'C'. We want an isometry f with f(A)=A', f(B)=B', f(C)=C'.
Construction of f (three simple moves):
1. Translate ABC so A goes to A'. Call the image A₁B₁C₁ (A₁ = A').
2. Rotate about A' so that the image of B₁ lies on the ray A'B'. (A rotation exists because AB = A'B' so we can place the segment of the same length along the same direction.)
3. If needed, reflect about the line A'B' to get the image of the second side to lie on the correct side of the ray (so that the orientation of the included angle matches). Allow reflection as part of the rigid motions.
After these moves we have an isometry f sending A to A' and sending the segment AB to the segment A'B'. Because rigid motions preserve angles, the image of ray AC is a ray at A' that makes the same angle with A'B' as AC made with AB. By the hypothesis ∠BAC = ∠B'A'C', that ray must be the ray A'C'. Because rigid motions also preserve distances and AC = A'C', the image of C lies on that ray at distance A'C' from A' — i.e. the image of C is exactly C'.
Therefore f(A)=A', f(B)=B', f(C)=C', so triangle ABC is carried exactly onto triangle A'B'C'. Hence the triangles are congruent. This is the SAS congruence theorem justified by rigid transformations.
Set-up. Let triangles ABC and A'B'C' satisfy AB = A'B', AC = A'C' and ∠BAC = ∠B'A'C'. We want an isometry f with f(A)=A', f(B)=B', f(C)=C'.
Construction of f (three simple moves):
1. Translate ABC so A goes to A'. Call the image A₁B₁C₁ (A₁ = A').
2. Rotate about A' so that the image of B₁ lies on the ray A'B'. (A rotation exists because AB = A'B' so we can place the segment of the same length along the same direction.)
3. If needed, reflect about the line A'B' to get the image of the second side to lie on the correct side of the ray (so that the orientation of the included angle matches). Allow reflection as part of the rigid motions.
After these moves we have an isometry f sending A to A' and sending the segment AB to the segment A'B'. Because rigid motions preserve angles, the image of ray AC is a ray at A' that makes the same angle with A'B' as AC made with AB. By the hypothesis ∠BAC = ∠B'A'C', that ray must be the ray A'C'. Because rigid motions also preserve distances and AC = A'C', the image of C lies on that ray at distance A'C' from A' — i.e. the image of C is exactly C'.
Therefore f(A)=A', f(B)=B', f(C)=C', so triangle ABC is carried exactly onto triangle A'B'C'. Hence the triangles are congruent. This is the SAS congruence theorem justified by rigid transformations.
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