The First Six Books of the Elements of Euclid

by Euclid; John Casey

2. If AB does not pass through the centre O, let fall the perpendicular OC on AB. Join OT, OB, OP. Then because OC, a line through the centre, cuts AB, which does not pass through the centre at right angles, it bisects it [iii.]. Hence, since AB is bisected in C and divided externally in P, the rectangle

		AP.BP + CB2	= CP2 [II. vi.];
	and	OC2	= OC2.

Hence, adding, since CB² + OC² = OB² [I. xlvii.], and CP² + OC² = OP², we get

	rectangle	AP.BP + OB2	= OP2;
	but	OT2 + PT2	= OP2 [I. xlvii.].
	Therefore	AP.BP + OB2	= OT2 + PT2;

and rejecting the equals OB² and OT², we have the rectangle

The two Propositions xxxv., xxxvi., may be included in one enunciation, as follows:—The rectangle AP.BP contained by the segments of any chord of a given circle passing through a fixed point P, either within or without the circle, is constant. For let O be the centre: join OA, OB, OP. Then OAB is an isosceles triangle, and OP is a line drawn from its vertex to a point P in the base, or base produced. Then the rectangle AP.BP is equal to the difference of the squares of OB and OP, and is therefore constant.

Cor. 1.—If two lines AB, CD produced meet in P, and if the rectangle AP.BP = CP.DP, the points A, B, C, D are concyclic (compare xxxv., Cor. 2).

Cor. 2.—Tangents to two circles from any point in their common chord are equal (compare xvii., Ex. 6).