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February 10th, 2015, 09:27 AM   #1
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Counterexample for Parallelogram

I just missed a math problem involving a justification of whether or not a quadrilateral must be a parallelogram. The quadrilateral had one pair of opposite sides that were congruent, and had one pair of congruent angles. Would this quadrilateral be a parallelogram? I can't think of any counterexample, and would like it if one of you could provide a counterexample.
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February 10th, 2015, 09:40 AM   #2
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It could be a trapezium (the congruent sides being non-parallel).
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February 10th, 2015, 09:44 AM   #3
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I suspect you could make an irregular quadrilateral if the congruent angles were at the two ends of one side.
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February 10th, 2015, 12:18 PM   #4
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This would be a reasonable counterexample, except that the congruent angles are also opposite from each other (I forgot to mention this).
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February 10th, 2015, 02:08 PM   #5
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Part way to solution. Add diagonal (not splitting the equal angles). The two triangles are "almost" congruent (SSA has two possibilities). Show that the alternate possibility can't hold.
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February 10th, 2015, 03:21 PM   #6
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Draw a parallelogram in which the known angle is acute and the added diagonal (that doesn't split the known angles) is shorter than the known side. This will make sure that the alternate possibility referred to above does exist and leads to the counterexample being sought.
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February 10th, 2015, 03:37 PM   #7
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This something I had in mind. I suspect that the quadrilateral is not convex.
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February 10th, 2015, 05:46 PM   #8
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It can be convex, but the construction I gave doesn't ensure that it's convex.
NonParallogram.PNG
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