To show that triangle \( \triangle ABC \) is congruent to triangle \( \triangle GHE \) through a sequence of transformations, we can describe the following steps:
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Translation: First, we need to translate \( \triangle ABC \) so that one of its vertices aligns with a vertex of \( \triangle GHE \).
We can translate point \( A' (1, -3) \) to point \( G (-2, 2) \). The translation vector can be calculated as: \[ (-2 - 1, 2 - (-3)) = (-3, 5) \] Thus, we translate triangle \( ABC \) by the vector \( (-3, 5) \):
- \( A' \) moves to \( G \) at \( (-2, 2) \)
- \( B (1, -1) \) translates to \( (-2, 4) \)
- \( C (4, -3) \) translates to \( (1, 2) \)
After this translation, the points will be:
- \( A'' = G (-2, 2) \)
- \( B' = (-2, 4) \)
- \( C' = (1, 2) \)
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Rotation: Next, we need to perform a rotation or reflection so that the angles and order of the points match. To align \( B' \) and \( E (-2, 5) \), we look at the configuration of the points:
- Notice that \( B' (-2, 4) \) is directly below \( E (-2, 5) \), suggesting a potential rotation around point \( G (-2, 2) \).
To transform \( B' \) to \( E \) and align the remaining points correctly, we can rotate \( 90^\circ \) counter-clockwise around point \( G (-2, 2) \):
- The coordinates will transform as follows:
- Rotating \( A' (-2, 2) \) does not change (it stays \( G \)).
- Rotating \( B' (-2, 4) \) around \( G \) results in \( E (-2, 5) \).
- Rotating \( C' (1, 2) \) around \( G (-2, 2) \) results in the new position that aligns with \( H (-4, 2) \).
With these transformations, we show:
- Translation of \( \triangle ABC \) such that point \( A' \) aligns with point \( G \).
- Rotation around \( G \) to align all corresponding points.
Therefore, \( \triangle ABC \) can be transformed into \( \triangle GHE \) through the described sequence of transformations, proving their congruence.