Dem.—If not, suppose another line OG drawn from the centre to be perpendicular to CD. Let OG cut the circle in F. Then because the angle OGC is right (hyp.) the angle OCG [I. xvii.] must be acute. Therefore [I. xix.] OC is greater than OG; but OC is equal to OF [I. Def. xxxii.]; therefore OF is greater than OG—that is, a part greater than the whole, which is impossible. Hence OC must be perpendicular to CD. Or thus: Since the perpendicular must be the shortest line from O to CD, and OC is evidently the shortest line; therefore OC must be perpendicular to CD. PROP. XIX.—Theorem. If a line (AB) be a tangent to a circle, the line (AC) drawn at right angles to it from the point of contact passes through the centre. If the centre be not in AC, let O be the centre. Join AO. Then because AB touches the circle, and OA is drawn from the centre to the point of contact, OA is at right angles to AB [xviii.]; therefore the angle OAB is right, and the angle CAB is right (hyp.); therefore OAB is equal to CAB—a part equal to the whole, which is impossible. Hence the centre must be in the line AC. Cor.—If a number of circles touch the same line at the same point, the locus of their centres is the perpendicular to the line at the point. Observation.—Propositions xvi., xviii., xix., are so related that any two can be inferred from the third by the “Rule of Identity.” Hence it would, in strict logic, be sufficient to prove any one of the three, and the others would follow. Again, these three theorems are limiting cases of Proposition i., Cor. 1., and Parts 1, 2, of Proposition iii., namely, when the points in which the chord cuts the circle become consecutive. PROP. XX.—Theorem. The angle (AOB) at the centre (O) of a circle is double the angle (ACB) at the circumference standing on the same arc. Dem.—Join CO, and produce it to E. Then because OA is equal to OC, the angle ACO is equal to OAC; but the angle AOE is equal to the sum of the two angles OAC, ACO. Hence the angle AOE is double the angle ACO. In like manner the angle EOB is double the angle OCB. Hence (by adding in figs. (α), (β), and subtracting in (γ)), the angle AOB is double of the angle ACB. Cor.—If AOB be a straight line, ACB will be a right angle—that is, the angle in a semicircle is a right angle (compare xxxi.). PROP. XXI.—Theorem. The angles (ACB, ADB) in the same segment of a circle are equal. Dem.—Let O be the centre. Join OA, OB. Then the angle AOB is double of the angle ACB [xx.], and also double of the angle ADB. Therefore the angle ACB is equal to the angle ADB. The following is the proof of the second part—that is, when the arc AB is not greater than a semicircle, without using angles greater than two right angles:—