Previous problems can be found here.
Part 1
5. In the diagram below, if triangle ABE = triangle and AEFC is drawn, then it could be proven that quadrilateral ABCD is a
(4) Parallelogram.
First: common sense. If the figure could be proved to be a rectangle or a rhombus, it would have to be a parallelogram as well. You can't have multiple answers. If it were a square, then all four would be true.
Second: figures are not necessarily drawn to scale (even if they don't tell you), so you shouldn't assume that it isn't a square based solely on the picture. Looking at an image isn't proof.
If the triangles are congruent, then by CPCTC (Corresponding parts of congruent triangles are congruent) we know that AB = CD. We also know, for the same reason, that angle BAE = DCF. Those two angles are alternate interior angles along the transversal. That makes AB || CD.
If two sides of a quadrilateral are both parallel and congruent, the shape is a parallelogram.
We do not have any additional information to prove (nor assume!) that the shape is a rhombus.
6. Under which transformation would triangle A'B'C', the image of triangle ABC, not be congruent to ABC?
(4) dilation with a scale factor of 2 centered at the origin.
A dilation increased the size, so it will no longer be congruent.
Continue to the next problems.
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