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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?

Michael Rozenberg
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Tom Finet
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  • Diagonals of a square are equal. All sides are equal. Diagonals bisect each other at $90^{\circ}$. – Paras Khosla Mar 26 '19 at 11:02
  • 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). – Paul Childs Mar 26 '19 at 11:39

1 Answers1

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Prove that $AC\perp BD$, $AC=BD$ and $AC$ and $BD$ have the same midpoint.

Michael Rozenberg
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    Should be AC = BD. You also need to have the converse re bisection, namely that BD bisects AC (they aren't equivalent). – Paul Childs Mar 26 '19 at 11:35
  • Thank you, @Paul ! I fixed. – Michael Rozenberg Mar 26 '19 at 11:56
  • Awesome. But don't forget the second part, else a kite may satisfy your conditions. AC must bisect BD and BD must bisect AC. – Paul Childs Mar 26 '19 at 12:05
  • It was my delirium. Thank you! I fixed again. – Michael Rozenberg Mar 26 '19 at 12:10
  • The solution seems to rely on assumption that $AC$ and $BD$ are diagonals. But they can appear to be parallel sides of the square... The test will fail. – user May 13 '20 at 14:32
  • @user We say about quadrilateral $ABCD$. See please better the starting problem. In this case $AC$ and $BD$ they are diagonals always by the definition of the quadrilateral. – Michael Rozenberg May 13 '20 at 15:22
  • OP starts with the sentence "I have points $A,B,C,D$". From this it is hard to conclude how the points are arranged. Nothing else follows from the definition of a general quadrilateral. I have no intention to compromise your solution. I just want to point out that a proof that given 4 points in general position form a square requires more steps. – user May 13 '20 at 15:36
  • @user We say about square $ABCD$. See please better the given. We know the ordering of our points. – Michael Rozenberg May 13 '20 at 15:37
  • Of course, if the variants that $ACBD$ or $ACDB$ are squares are not acceptable, your solution is perfect. – user May 13 '20 at 15:43
  • @user they are not acceptable by the definition of the quadrilateral. – Michael Rozenberg May 13 '20 at 15:45
  • I would say by the definition of the square. A general quadrilateral can have self-intersection. – user May 13 '20 at 15:47
  • @user Maybe, but it's not in the school. – Michael Rozenberg May 13 '20 at 15:50