Question

Quad WXYZ must be a special figure to meet the conditions stated. Write the best name for that special quadrilateral.

Given that $\angle 1\cong \angle 2\cong \angle 3\cong \angle 4$. Prove that $EFGH$is a rhombus.

Short answer

The parallelogram is a rhombus.

Step-by-step solution

Step 1. Check the figure.

Consider the figure.

Step 2. Step description.

From the figure, for the lines $\overline{HE}$and $\overline{GF}$and the transversal $\overline{EG}$, the corresponding angles $\angle 1$and $\angle 3$are congruent.

So, the lines $\overline{HE}$and $\overline{GF}$are parallel.

Similarly, for the lines $\overline{HG}$and $\overline{EF}$and the transversal $\overline{EG}$, the corresponding angles $\angle 2$and $\angle 4$are congruent.

So, the lines $\overline{HG}$and $\overline{EF}$are parallel.

Consider the quadrilateral $EFGH$.

It is known that if pairs of opposite sides of any quadrilateral are parallel, then it is a parallelogram.

Step 3. Step description.

Here,$\overline{HE}\parallel \overline{GF}$ and$\overline{HG}\parallel \overline{EF}$ so,$EFGH$ is a parallelogram.

Now, for the triangle $△EGF$, $\angle 2\cong \angle 3$.

So, the sides$\overline{GF}$ and$\overline{EF}$ are congruent.

As $\overline{EF}\cong \overline{GF}$, thus the parallelogram$EFGH$ is a rhombus.