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A followup to my previous question -

Assume we have rhombus wxyz. If I draw a line from w to y (opposite vertices), this would bisect angles w and y and create two triangles, wxy and wzy. If I show that the triangles overlap perfectly when reflected over wy, does that imply wxyz is a square? I believe it does because if angle z or x are not 90, then they would not overlap. And if one angle of a rhombus is 90, then all are and it is a square.

NOTE: I am NOT asking about congruence. I realize that the triangles would be congruent. I am asking if they would perfectly overlap by only reflection if and only if the rhombus was a square.

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    $\begingroup$ Draw a a "diamond" which is thinner than it is tall. Kind of like this $\diamondsuit$, but less fat. $\endgroup$ – André Nicolas Sep 26 '13 at 3:26
  • $\begingroup$ The intuitive idea underlying congruence is "perfectly overlap" if properly aligned. If they overlap by reflection, then they are congruent. $\endgroup$ – A.Ellett Sep 26 '13 at 3:47
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There are several important properties of rhombi:

  • Their diagonals bisect the angles of the rhombus
  • Their diagonals bisect each other
  • Their diagonals are perpendicular

Since the sides of a rhombus are congruent, all a single diagonal will do is divide the rhombus into two congruent isosceles triangles. In other words, triangle WXY will be congruent to WZY for any rhombus WXYZ. Similar, triangle XYZ and triangle XWZ are also congruent.

However, if you can show that triangles WXY and XWZ are congruent, then you do indeed have a square.

If you can show that the diagonals are congruent, then you can prove that you have a square.

Here's a diagram to illustrate the rotation about the diagonal.

enter image description here

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  • $\begingroup$ I realize that the triangles would be congruent, but that is not my question. I'm asking specifically about the case when they overlap via just reflection (as opposed to rotation/translation). I believe that if the rhombus is not a square then you can draw diagonals such that the triangle reflections will NOT perfectly overlap (even though they are still congruent). $\endgroup$ – David says Reinstate Monica Sep 26 '13 at 3:25
  • $\begingroup$ @Dgrin91 My response is about "reflection" about the diagonal. You are mistaken about the triangles on either side of a diagonal. $\endgroup$ – A.Ellett Sep 26 '13 at 3:29
  • $\begingroup$ Lets assume a rhombus which is not a square though. If we draw a diagonal and reflect one of the congruent triangles about the diagonal, then it would not perfectly overlap the other triangle even though they are congruent. I believe this is because the corresponding sides need to be flipped. $\endgroup$ – David says Reinstate Monica Sep 26 '13 at 3:35
  • $\begingroup$ @Dgrin91 I think you are not understanding correctly what it means to "reflect" around something. In a rhombus, to reflect about a diagonal would be like folding the rhombus along its diagonal. If it is indeed a rhombus, the two triangles will match perfectly. $\endgroup$ – A.Ellett Sep 26 '13 at 3:37
  • $\begingroup$ Yeah, that is what I am thinking. Im just guessing at my reflection theory, so I could be (and it seems am) totally wrong. I haven't actually tried to do this with a real rhombus. $\endgroup$ – David says Reinstate Monica Sep 26 '13 at 3:38

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