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I have points $A, B, C, D$ and I need to show that $ABCD$ is a square. I tried this by showing all four sides have equal length and that two angles sharing a side are $90^{\circ}$ degrees. Is there a more concise way to prove the points form a square?

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  • $\begingroup$ Diagonals of a square are equal. All sides are equal. Diagonals bisect each other at $90^{\circ}$. $\endgroup$ – Paras Khosla Mar 26 at 11:02
  • $\begingroup$ One point finer than yours is that if the shared side is BC you only have to show AB BC and CD are equal (unless the shape can be self intersecting). $\endgroup$ – Paul Childs Mar 26 at 11:39
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Prove that $AC\perp BD$, $AC=BD$ and $AC$ and $BD$ have the same midpoint.

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    $\begingroup$ Should be AC = BD. You also need to have the converse re bisection, namely that BD bisects AC (they aren't equivalent). $\endgroup$ – Paul Childs Mar 26 at 11:35
  • $\begingroup$ Thank you, @Paul ! I fixed. $\endgroup$ – Michael Rozenberg Mar 26 at 11:56
  • $\begingroup$ Awesome. But don't forget the second part, else a kite may satisfy your conditions. AC must bisect BD and BD must bisect AC. $\endgroup$ – Paul Childs Mar 26 at 12:05
  • $\begingroup$ It was my delirium. Thank you! I fixed again. $\endgroup$ – Michael Rozenberg Mar 26 at 12:10

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