(2). Let the arc ACE be greater than a semicircle. Join CE. Then the angle ACE is evidently less than ACB; but ACB is right; therefore ACE is acute. (3). Let the arc ACD be less than a semicircle; then evidently, from (1), the angle ACD is obtuse. Cor. 1.—If a parallelogram be inscribed in a circle, its diagonals intersect at the centre of the circle. 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. Or by the method of limits, see Townsend’s Modern Geometry, vol. i., page 14. The angle BAC is equal to BDC [xxi.]. Now let the point B move until it becomes consecutive to A; then AB will be a tangent, and BD will coincide with AD, and the angle BDC with ADC. Hence, if AX be a tangent at A, AC any chord, the angle which the tangent makes with the chord is equal to the angle in the alternate segment. Exercises. 1. If two circles touch, any line drawn through the point of contact will cut off similar segments. 2. If two circles touch, and any two lines be drawn through the point of contact, cutting both circles again, the chord connecting their points of intersection with one circle is parallel to the chord connecting their points of intersection with the other circle.