Question

Given $▱AZBY;\overline{ZY}\cong \overline{BX};\angle 1\cong \angle 2$. Prove $AZBY$is a rhombus.

Short answer

Therefore, AZBY is a rhombus.

Step-by-step solution

Step 1. Check the figure.

Consider the figure.

Step 2. Step description.

The CPCTC theorem states that when two triangles are congruent, then every corresponding part of one triangle is congruent to the other.

Step 3. Step description.

Consider that $\overline{ZY}\cong \overline{BX};\angle 1\cong \angle 2$.

Since the vertical angles are congruent thus $\angle AXB\cong \angle YXZ$.

By the angle-angle-side postulate $\Delta AXB\cong \Delta YXZ$.

By the CPCTC theorem, this implies that $AX\cong XZ;BX\cong XY$.

Therefore, $\angle YAB$is a right angle by the definition of a parallelogram.

From the figure, $\overline{AZ}$is perpendicular bisector to $\overline{BY}$and by the definition of rhombus, $\overline{BY}$is perpendicular bisector to $\overline{AZ}$.

Therefore, ABZY is a rhombus.