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Consider rectangle EFGH. Prove EG is congruent to HF

User Lubomira
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To prove that EG is congruent to HF in rectangle EFGH, we can use the following argument:

Recall that in a rectangle, opposite sides are parallel and congruent. Since EF is parallel to GH and congruent to GH, and EG is a diagonal of the rectangle, we can use the Pythagorean theorem to show that EG is congruent to HF.

In particular, let x be the length of EG. Then, by the Pythagorean theorem, we have:

x^2 = EF^2 + FG^2

Since EF is congruent to GH and FG is congruent to HE (since they are opposite sides of a rectangle), we have:

x^2 = GH^2 + HE^2

By the Pythagorean theorem again, we have:

x^2 = HF^2

Therefore, x = HF, which means that EG is congruent to HF in rectangle EFGH.
User Vivek Srinivasan
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