Does a rhombus bisector creating two equal triangles by reflection imply a square?

Question

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.

Answer 1

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.