The First Six Books of the Elements of Euclid

by Euclid; John Casey

3. Let AB pass through the centre, and cut CD, which does not pass through the centre obliquely. Let O be the centre. Draw OF perpendicular to CD [I. xi.]. Join OC, OD. Then, since CD is cut at right angles by OF, which passes through the centre, it is bisected in F [iii.], and divided unequally in E. Hence

		CE.ED + FE2	= FD2 [II. v.],
	and	OF2	= OF2.

Hence, adding, since FE² + OF² = OE² [I. xlvii.], and FD² + OF² = OD², we get

Again, since AB is bisected in O and divided unequally in E,

		AE.EB + OE2	= OB2 [II. v.].
	Therefore	CE.ED + OE2	= AE.EB + OE2.
	Hence	CE.ED	= AE.EB.

4. Let neither chord pass through the centre. Through the point E, where they intersect, draw the diameter FG. Then by 3, the rectangle FE.EG is equal to the rectangle AE.EB, and also to the rectangle CE.ED. Hence the rectangle AE.EB is equal to the rectangle CE.ED.

Cor. 1.—If a chord of a circle be divided in any point within the circle, the rectangle contained by its segments is equal to the difference between the square of the radius and the square of the line drawn from the centre to the point of section.

Cor. 2.—If the rectangle contained by the segments of one of two intersecting lines be equal to the rectangle contained by the segments of the other, the four extremities are concyclic.