Asked by paige
Describe a sequence of rigid transformations that could be used to show that △ABC≅△EFD
. Be sure to explain how you know that each pair of corresponding vertices will overlap perfectly. Then, explain why this sequence illustrates the SAS Congruence Theorem.
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Answered by
GPT-5 mini
AI
I will give a concrete 3-step rigid-motion sequence that carries △ABC onto △EFD, then explain why each corresponding vertex lands exactly on its match and how this is an example of SAS.
Assume you are given that AB = EF, AC = ED, and ∠BAC = ∠FED (these are the SAS data). The rigid-motion sequence:
1. Translation: translate △ABC so that A goes to E.
- Call the translated triangle A'B'C' (so A' = E). Translations preserve all distances and angles, so B' is at the same distance from E as B was from A (|EB'| = |AB|) and C' is at the same distance from E as C was from A (|EC'| = |AC|).
2. Rotation about E: rotate the translated triangle A'B'C' around point E by the angle that takes the ray EA' (which is EA, now) onto the ray EF.
- Rotations preserve distances and angles. After this rotation the image of B' lies somewhere on the ray EF at distance |AB| from E. Because AB = EF, that image point must be exactly F (there is only one point on ray EF at distance EF from E). Thus B is carried to F. Call the image of C under this rotation C''.
3. (If needed) Reflection across line EF: if after the rotation C'' coincides with D, you are done. If not — equivalently, if the triangles had opposite orientation so that a rotation could not place C on D — reflect the rotated image across the line EF. A reflection is also rigid, so it preserves lengths and angles; reflecting will send the already-matched points E and F to themselves and will carry C'' to the unique point on the other side of EF at distance |AC| = |ED|, which must be D. After this reflection every vertex matches: A → E, B → F, C → D.
Why each vertex overlaps perfectly
- Translations, rotations and reflections are rigid motions: they preserve all segment lengths and angle measures. At each step we use uniqueness of a point at a given distance on a given ray or line to force coincidence. For example, after we translate and rotate so the ray through E that should contain the image of B is EF, the image of B must be the point on EF at distance AB from E. Because AB = EF, that unique point is F, so B overlaps F. The same uniqueness argument (point at distance AC from E on the appropriate ray) forces C to overlap D.
Why this illustrates SAS
- The sequence used only rigid motions, so it carried two sides AB and AC and the included angle ∠BAC to EF, ED and ∠FED respectively. Since lengths AB = EF and AC = ED and the included angles are equal, the rigid motions place the two sides and the included angle of △ABC exactly onto the corresponding two sides and included angle of △EFD. Thus two sides and their included angle coincide under the motion, which is precisely the condition of SAS: if two triangles have two pairs of corresponding sides equal and the included angles equal, the triangles are congruent. The rigid-motion construction produces the explicit congruence mapping that SAS guarantees.
Assume you are given that AB = EF, AC = ED, and ∠BAC = ∠FED (these are the SAS data). The rigid-motion sequence:
1. Translation: translate △ABC so that A goes to E.
- Call the translated triangle A'B'C' (so A' = E). Translations preserve all distances and angles, so B' is at the same distance from E as B was from A (|EB'| = |AB|) and C' is at the same distance from E as C was from A (|EC'| = |AC|).
2. Rotation about E: rotate the translated triangle A'B'C' around point E by the angle that takes the ray EA' (which is EA, now) onto the ray EF.
- Rotations preserve distances and angles. After this rotation the image of B' lies somewhere on the ray EF at distance |AB| from E. Because AB = EF, that image point must be exactly F (there is only one point on ray EF at distance EF from E). Thus B is carried to F. Call the image of C under this rotation C''.
3. (If needed) Reflection across line EF: if after the rotation C'' coincides with D, you are done. If not — equivalently, if the triangles had opposite orientation so that a rotation could not place C on D — reflect the rotated image across the line EF. A reflection is also rigid, so it preserves lengths and angles; reflecting will send the already-matched points E and F to themselves and will carry C'' to the unique point on the other side of EF at distance |AC| = |ED|, which must be D. After this reflection every vertex matches: A → E, B → F, C → D.
Why each vertex overlaps perfectly
- Translations, rotations and reflections are rigid motions: they preserve all segment lengths and angle measures. At each step we use uniqueness of a point at a given distance on a given ray or line to force coincidence. For example, after we translate and rotate so the ray through E that should contain the image of B is EF, the image of B must be the point on EF at distance AB from E. Because AB = EF, that unique point is F, so B overlaps F. The same uniqueness argument (point at distance AC from E on the appropriate ray) forces C to overlap D.
Why this illustrates SAS
- The sequence used only rigid motions, so it carried two sides AB and AC and the included angle ∠BAC to EF, ED and ∠FED respectively. Since lengths AB = EF and AC = ED and the included angles are equal, the rigid motions place the two sides and the included angle of △ABC exactly onto the corresponding two sides and included angle of △EFD. Thus two sides and their included angle coincide under the motion, which is precisely the condition of SAS: if two triangles have two pairs of corresponding sides equal and the included angles equal, the triangles are congruent. The rigid-motion construction produces the explicit congruence mapping that SAS guarantees.
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