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.