To prove that triangles \( \triangle ABC \) and \( \triangle GHE \) are congruent, we can describe a sequence of transformations that will map \( \triangle ABC \) onto \( \triangle GHE \).
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Translation: First, we need to move \( \triangle ABC \) so that point \( A \) overlaps with point \( G \). The coordinates of point \( A \) are \( (1, -3) \) and the coordinates of point \( G \) are \( (-2, 2) \).
To translate point \( A(1, -3) \) to point \( G(-2, 2) \): \[ \text{Translate } A \text{ by } \left(-2 - 1, 2 + 3\right) = \left(-3, 5\right). \] So, we need to apply the translation vector \( (-3, 5) \).
The translation can be expressed as: \[ (x, y) \rightarrow (x - 3, y + 5). \]
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Rotation: After the translation, we have the new positions of the points of \( \triangle ABC \):
- \( A' = G(-2, 2) \)
- \( B' = A'(1, -1) - (3, 5) = (1-3, -1+5) = (-2, 4) \)
- \( C' = A'(4, -3) - (3, 5) = (4-3, -3+5) = (1, 2) \)
Now, we have the new \( A', B', C' \) as \( (-2, 2), (-2, 4), (1, 2) \).
We need to rotate the triangle \( ABC \) (now at these new positions) so that it matches \( \triangle GHE \). It looks like we need to rotate \( 90^\circ \) clockwise around point \( G(-2,2) \):
For a point \( (x, y) \):
- The 90-degree clockwise rotation transformation can be represented with the center \( (-2, 2) \): \[ (x', y') = (hx - k + k, -hy + k + k), \] where \( (k,r) \) is the center of rotation \( G \) which is \( (-2, 2) \).
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Mapping: Applying the transformation:
- \( B'(-2,4) \) will go to \( H(-4, 2) \)
- \( C'(1,2) \) will go to \( E(-2,5) \)
Thus, through these transformations, we have moved \( \triangle ABC \) to congruently match \( \triangle GHE \).
Summary of Transformations:
- Translate \( \triangle ABC \) by \( (-3, 5) \).
- Rotate the resulting triangle \( 90^\circ \) clockwise about point \( G(-2, 2) \).
This proves the congruence of \( \triangle ABC \) and \( \triangle GHE \) through these transformations.
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