Question

prove that if a parallelogram is a rhombus, then its diagonals are perpendicular bisectors of each other.
given: wxyz is a rhombus.
prove: wy and xz are perpendicular bisectors of each other.
complete the proof below.
statement
wxyz is a rhombus
wx ≅ xy ≅ yz ≅ zw
reason
given

Explanation:

Step1: Properties of a rhombus

A rhombus is a parallelogram with all sides equal. So, if \(WXYZ\) is a rhombus, then \(WX\cong XY\cong YZ\cong ZW\) by the definition of a rhombus.

Step2: Diagonals of a rhombus

Let the diagonals \(WY\) and \(XZ\) intersect at point \(P\). In \(\triangle WPZ\) and \(\triangle WPX\), \(WZ = WX\) (sides of a rhombus), \(PZ=PX\) (diagonals of a parallelogram bisect each other, and a rhombus is a parallelogram), and \(WP = WP\) (common side). By SSS (Side - Side - Side) congruence criterion, \(\triangle WPZ\cong\triangle WPX\). Then \(\angle WPZ=\angle WPX\). Since \(\angle WPZ+\angle WPX = 180^{\circ}\) (linear pair), \(\angle WPZ=\angle WPX = 90^{\circ}\). So, \(WY\perp XZ\). Also, since \(WXYZ\) is a parallelogram, the diagonals bisect each other.

Answer:

The diagonals of a rhombus are perpendicular bisectors of each other. So, if \(WXYZ\) is a rhombus, then \(WY\) and \(XZ\) are perpendicular bisectors of each other. And by the definition of a rhombus, \(WX\cong XY\cong YZ\cong ZW\)