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How do I prove if EFGH is a rhombus in this paragraph proof?

How Do I Prove If EFGH Is A Rhombus In This Paragraph Proof class=

Sagot :

sqdancefan

Explanation:

There are many approaches you can take. Perhaps one of the easiest is to show the sides of the rhombus are all the same length. This can be done several ways. Here are a couple of them.

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1.

A square has degree 4 rotational symmetry, so will map to itself when rotated by any multiple of 90°. This means the midpoints of adjacent sides will map to themselves, and any segment joining the midpoints of adjacent sides will map to itself. When segments map to themselves, they are congruent, hence the four sides of EFGH are congruent, and the figure is a rhombus.

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2.

The sides of a square are the same length and all corner angles are right angles. Midpoints of the sides divide the sides into congruent parts, so all of the segments PF, FQ, QG, GR, and so on are congruent. This means all of the triangles PFE, QGF, RHG, and SEH are congruent by the LL postulate. Corresponding sides of congruent triangles are congruent, so the sides of EFGH are congruent, and the figure is a rhombus.

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