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I have proven that a quadrilateral is a parallelogram, and that it is also a rhombus. Does this imply that it is ultimately a square?

I believe it does because all sides are equal, and opposite sides are parallel, and I believe the only way this is possible is if the angles are all 90, but I'm not sure if that is true. +1 if you can link to a proof that says that (or prove it yourself).

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  • $\begingroup$ You probably mean "rectangle and rhombus = square". $\endgroup$ – lhf Sep 26 '13 at 3:06
  • $\begingroup$ @lhf No, I meant parallelagram and rhombus. I was just wrong. $\endgroup$ – David Grinberg Sep 26 '13 at 3:13
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Every rhombus is a parallelogram and a rhombus with right angles is a square. Hence a quadrilateral which is a rhombus is a parallelogram, but not necessarily a square.

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  • $\begingroup$ Whoops, forgot rhombus has to be parallel. You are correct, thanks. $\endgroup$ – David Grinberg Sep 26 '13 at 3:04
  • $\begingroup$ No problems @ Dgrin91 $\endgroup$ – George Tomlinson Sep 26 '13 at 3:05
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    $\begingroup$ As a side note, Euclid's original definition of rhombus actually excludes a square ... so you have to be careful. $\endgroup$ – Calvin Lin Sep 26 '13 at 4:01
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A rhombus is always a parallelogram. In order to guarantee that you have a square, you'd have to show that all four angles are congruent.

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