The First Six Books of the Elements of Euclid

by Euclid; John Casey

In an obtuse-angled triangle (ABC), the square on the side (AB) subtending the obtuse angle exceeds the sum of the squares on the sides (BC, CA) containing the obtuse angle, by twice the rectangle contained by either of them (BC), and its continuation (CD) to meet a perpendicular (AD) on it from the opposite angle.

Dem.—Because BD is divided into two parts in C, we have

		BD2 = BC2	+ CD2 + 2BC.CD [iv.]
	and		AD2 = AD2.

Hence, adding, since [I. xlvii.] BD² + AD² = AB², and CD² + AD² = CA², we get

Therefore AB² is greater than BC² + CA² by 2BC.CD.

The foregoing proof differs from Euclid’s only in the use of symbols. I have found by experience that pupils more readily understand it than any other method.

Or thus: By the First Book: Describe squares on the three sides. Draw AE, BF, CG perpendicular to the sides of the squares. Then it can be proved exactly as in the demonstration of [I. xlvii.], that the rectangle BG is equal to BE, AG to AF, and CE to CF. Hence the sum of the two squares on AC, CB is less than the square on AB by twice the rectangle CE; that is, by twice the rectangle BC.CD.