On the side BC of rectangle ABCD, for which AB = 24 and AD = 31, point E is marked so that ∠EAB = 45 °. Find ED.

In the rectangle ABCD AB = CD = 24 and AD = BC = 31.

Consider triangle ABE. In it ∠ ABE = 90º, ∠ EAB = 45º, so ∠ BEA = 180 – 90 – 45 = 45º. Therefore, ∆ ABE is isosceles, and AB = BE = 24.

EC = BC – BE = 31 – 24 = 7.

Now consider the triangle ECD. In it, ∠ ECD is straight, therefore, ∆ ECD is rectangular with hypotenuse ED. By the Pythagorean theorem, we find ED:

ED = √ (EC2 + CD2) = √ (72 + 242) = √ (49 + 576) = √625 = 25.