Chapter 5.4, Problem 28WE

To determine

To show: $ABZY$ is a rhombus.

Answer to Problem 28WE

  $ABZY$ is a rhombus.

Explanation of Solution

Given information: $ABZY$ is a parallelogram,

  $\overline{ZY}\cong \overline{BX}$

  $\angle 1\cong \angle 2$

  

Proof:

Statement   $\overline{ZY}\cong \overline{BX}$   $\angle 1\cong \angle 2$   $\angle AXB\cong \angle YXZ$   $\Delta AXB\cong \Delta YXZ$   $AX\cong XZ$   $BX\cong XY$ Hence $\angle YAB$ is a right angle$\overline{AZ}$ is perpendicular bisector to $\overline{BY}$and <$\overline{BY}$ is perpendicular bisector to $\overline{AZ}$ Therefore, $ABZY$is a rhombus

Reason Given Given Vertical angles are congruent AAS CPCTC CPCTC Definition of parallelogram Definition of a perpendicular bisector Definition of a rhombus