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Let quadrilateral n have vertices A, B, C, and D. If side AB is congruent to side CD, and angles ABC and CDA are also congruent. Can you show this is not necessarily a parallelogram?

It does not follow the conditions for a parallelogram, yet my math teacher and I were unable to show a counter example or prove that it is or is not.

One cannot dissect it and use triangle congruence, for it leads to ASS.

Can someone show it is not a parallelogram, or even better, show a counter example?

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1 Answer

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Yes, it can be a non-parallelogram.

Regarding the triangles $\triangle ABC$ and $\triangle CDA$ - due to the two solutions for the inverse sine, $\measuredangle DAC$ and $\measuredangle ACB$ can be different.

Note that if the given $\measuredangle ABC = \measuredangle CDA$ is obtuse, there is only one viable solution for the inverse sine and the shape is indeed a parallelogram.

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