Chapter 5.5, Problem 7ST2

To determine

To prove: $EFGH$ is a rhombus

Explanation of Solution

Given:

The diagram:

  $\angle 1\cong \angle 2\cong \angle 3\cong \angle 4$

Consider the figure shown below:

It is given that, for the lines $\overline{HE}$ and $\overline{GF}$ and the transversal $\overline{EG}$ , the corresponding angles $\angle 1$ and $\angle 3$ are congruent.

So, by theorem 3-5, 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, by theorem 3-5, the lines $\overline{HG}$ and $\overline{EF}$ are parallel.

Consider the quadrilateral $EFGH$ .

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

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

Now, it is given that for the triangle $△EGF$ , $\angle 2\cong \angle 3$

So, by theorem 4-2, the sides $\overline{GF}$ and $\overline{EF}$ are congruent.

As $\overline{EF}\cong \overline{GF}$ , so by theorem 5-17, the parallelogram $EFGH$ is a rhombus.