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91Ó°ÊÓ

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

Given that ∠1≅∠2≅∠3≅∠4. Prove that EFGHis a rhombus.

Short Answer

Expert verified

The parallelogram is a rhombus.

Step by step solution

01

Step 1. Check the figure.

Consider the figure.

02

Step 2. Step description.

From the figure, for the lines HE¯and GF¯and the transversal EG¯, the corresponding angles ∠1and ∠3are congruent.

So, the lines HE¯and GF¯are parallel.

Similarly, for the lines HG¯and EF¯and the transversal EG¯, the corresponding angles ∠2and ∠4are congruent.

So, the lines HG¯and 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.

03

Step 3. Step description.

Here,HE¯∥GF¯ andHG¯∥EF¯ so,EFGH is a parallelogram.

Now, for the triangle △EGF, ∠2≅∠3.

So, the sidesGF¯ andEF¯ are congruent.

As EF¯≅GF¯, thus the parallelogramEFGH is a rhombus.

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