The First Six Books of the Elements of Euclid

by Euclid; John Casey

6. If the sum of two arcs, AC, CB of a circle be less than a semicircle, the rectangle AC.CB contained by their chords is equal to the rectangle contained by the radius, and the excess of the chord of the supplement of their difference above the chord of the supplement of their sum.—Catalan.

Dem.—Draw DE, the diameter which is perpendicular to AB, and draw the chords CF, BG parallel to DE. Now it is evident that the difference between the arcs AC, CB is equal to 2CD, and therefore = CD + EF. Hence the arc CBF is the supplement of the difference, and CF is the chord of that supplement. Again, since the angle ABG is right, the arc ABG is a semicircle. Hence BG is the supplement of the sum of the arcs AC, CB; therefore the line BG is the chord of the supplement of the sum. Now (Ex. 1), the rectangle AC.CB is equal to the rectangle contained by the diameter and CI, and therefore equal to the rectangle contained by the radius and 2CI; but the difference between CF and BG is evidently equal to 2CI. Hence the rectangle AC.CB is equal to the rectangle contained by the radius and the difference between the chords CF, BG.

7. If we join AF, BF we find, as before, the rectangle AF.FB equal to the rectangle contained by the radius and 2FI—that is, equal to the rectangle contained by the radius and the sum of CF and BG. Hence—If the sum of two arcs of a circle be greater than a semicircle, the rectangle contained by their chords is equal to the rectangle contained by the radius, and the sum of the chords of the supplements of their sum and their difference.

8. Through a given point draw a transversal cutting two lines given in position, so that the rectangle contained by the segments intercepted between it and the line may be given.

PROP. XXXVI.—Theorem.

If from any point (P) without a circle two lines be drawn to it, one of which (PT) is a tangent, and the other (PA) a secant, the rectangle (AP, BP) contained by the segments of the secant is equal to the square of the tangent.

Dem.—1. Let PA pass through the centre O. Join OT. Then because AB is bisected in O and divided externally in P, the rectangle AP.BP + OB² is equal to OP² [II. vi.]. But since PT is a tangent, and OT drawn from the centre to the point of contact, the angle OTP is right [xviii.]. Hence OT² + PT² is equal to OP².

	Therefore	AP.BP + OB2	= OT2 + PT2;
	but	OB2	= OT2.
		Hence the rectangle AP.BP	= PT2.