The First Six Books of the Elements of Euclid

by Euclid; John Casey

Cor. 2.—Find the centre of a circle by means of a carpenter’s square.

Cor. 3.—From a point outside a circle draw two tangents to the circle.

PROP. XXXII.—Theorem.

If a line (EF) be a tangent to a circle, and from the point of contact (A) a chord (AC) be drawn cutting the circle, the angles made by this line with the tangent are respectively equal to the angles in the alternate segments of the circle.

Dem.—(1). If the chord passes through the centre, the Proposition is evident, for the angles are right angles; but if not, from the point of contact A draw AB at right angles to the tangent. Join BC. Then because EF is a tangent to the circle, and AB is drawn from the point of contact perpendicular to EF, AB passes through this centre [xix.]. Therefore the angle ACB is right [xxxi.]. Hence the sum of the two remaining angles ABC, CAB is one right angle; but the angle BAF is right (const.); therefore the sum of the angles ABC, BAC is equal to BAF. Reject BAC, which is common, and we get the angle ABC equal to the angle FAC.

(2). Take any point D in the arc AC. It is required to prove that the angle CAE is equal to CDA.

Since the quadrilateral ABCD is cyclic, the sum of the opposite angles ABC, CDA is two right angles [xxii.], and therefore equal to the sum of the angles FAC, CAE; but the angles ABC, FAC are equal (1). Reject them, and we get the angle CDA equal to CAE.

Or thus: Take any point G in the semicircle AGB. Join AG, GB, GC. Then the angle AGB = FAB, each being right, and CGB = CAB [xxi.]. Therefore the remaining angle AGC = FAC. Again, join BD, CD. The angle BDA = BAE, each being right, and CDB = CAB [xxi.]. Hence the angle CDA = CAE.—Lardner.