Chapter 4.4, Problem 13PSB

To determine

To Prove: WXYZ is a rhombus.

Explanation of Solution

Given information: WY and XZ are perpendicular bisectors of each other.

Formula used: congruency of triangle.

Proof:

  

  $\begin{array}{l}WY\text{and}XZ\text{bisect}\text{each}\text{other}\text{at}{90}^{\circ }(\text{given})\\ \therefore WO=OY,OX=OZ.\\ \angle WOX=\angle YOX(\text{each}\text{equals }{90}^{\circ })\\ OX=OX(\text{reflection}\text{property})\\ \therefore \Delta XOW\cong \Delta XOY(\text{SAS})\\ \therefore WX=XY(\text{CPCT})\\ \text{Similarly}XY=YZ\text{and}YZ=WZ.\\ \therefore WXYZ\text{is}\text{a}\text{Rhombus}\text{.}\end{array}$