68.7k views
3 votes
Consider rectangle EFGH. Prove EG is congruent to HF

User Lubomira
by
8.6k points

1 Answer

1 vote
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
by
8.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.