A a C'; draw CD at right angles to AB, and meeting the circumference in D; produce DC to meet the circumference again in E. Then bisect DE in 0; and since DE is a diameter (49 Cor.), :O is the centre of the circle. 51. PROP. III. To describe a circle whose circumference shall pass through three given points. Let A, B, C, be the three given points; join AB, and BC. Bisect AB in D, and BC in E; from D draw DF at right angles to AB; and from E draw EG at right angles to BC, intersecting DF in 0; with centre 0 and radius 0 A describe a circle, and its circumference shall pass through A, B, and C. Join A0, BO, CO; then in the triangles ADO, BDO, AD=BD, .. AD, DO are equal to BD, DO, each to each, and <AD0=- BDO, .. A0= BO (24). In the same way it may be shewn that BO = CO; .. AO= BO=CO; that is, a circle described with centre O and radius A0, will pass through the points A, B, C. N. B. If the given points A, B, C be in one and the same straight line, this construction will fail, because then DF and EG being at right angles to the same straight line will be parallel to each other, and never meet at all in 0. In this particular case there is no circle whose circumference can be made to pass through the three given points*. Also, if it be required to describe a circle whose circumference shall pass through two given points, A, B, it is plain, that there will be an infinite number of such circles, having their centres in the indefinite straight line DF. 52. PROP. IV. The angle which any arc of a circle subtends at the centre of the circle is double of the angle which it subtends at the circumferencet. * From this it follows that no straight line can meet the circumference of a circle in more than two points. + The angle which an arc, or other magnitude, subtends at a given a = = S Let AB be any arc of a circle whose centre is 0; C any point in the other portion of the circumference. Join AO, BO, AC, BC; then <AOB, which the arc AB subtends at 0, shall be double of -ACB which it subtends at C. Join Co, and produce it to meet the circumference in D; then since OA = OC, ZOAC= 2 ОСА (26); and LAOD = -OAC + < OCA (37) = twice GOCA. Similarly - BOD = twice z OCB; .. - AOB = -A0D+ - BOD (22), = twice OCA + twice <OCB, =twice ACB. If the point C be taken so that the centre 0 falls without the ACB, the construction is the same, and the proof also, except that angles are subtracted instead of added (22). Cor. Since 2AOB in the same circle is always the same for a given arc AB, it follows that <ACB, which is half of 2AOB, is the same whatever point C in the circumference be taken ; that is, if E be any other point in the circumference, and AE, BE be joined, - AEB = - ACB. These latter angles are, for shortness, called angles • in a segment'; and thus, with Euclid, we say, all angles in the same segment are equal to one another'. It is important for the student to make himself quite sure of the meaning of the phrase'angle in a segment'. In the 1st place, a segment, as already defined, is a portion of a circle bounded by an arc and the chord of the arc. Then an 'angle in the segment' is the angle which that chord subtends at any point in the arc. 53. PROP. V. In any four-sided rectilineal figure, which has all its angular points in the circumference of the same circle *, each pair of opposite angles is equal to two right angles. point, means the angle formed by two straight lines joining that point and the extreme points of the arc, or other magnitude. * This is what is meant by the expression, sometimes used, 'a quadrilateral inscribed in a circle'. To be inscribed it is necessary, that all the angular points fall upon the circumference of the circle. Let ABCD be a four-sided plane figure, having its angular points A, B, C, D, in the circumference of a circle; then - ABC + 2 ADC=two right angles; and likewise 2 BAD+2 BCD = two right angles. Join AC, BD; then in the triangle ABC, ABC + BAO + ZACB = two rightangles (37); but by (52) - BAC = 2 BDC, being angles “in the same segment' BADC. Also ZACB = LADB, being angles in the same segment' ADCB;: -ABC +2BDC + ZADB= two right angles; and 2 BDC + ZADB = 2 KDC (99), 1.4 ABC + 4 ADC = two right angles. In the same manner it may be shewn that < BAD+ < BCD = two right angles. COR. Hence no parallelogram except a rectangle can be“ inscribed in a circle ; because the opposite angles in every parallelogram are equal to one another; and therefore in this case each of them must be a right angle. 54. Prop. VI. The 'angle in a segment' equal to a semi-circle is a right angle ; in a segment greater than a semi-circle is less than a right angle ; and in a segment less than a semi-circle is greater than a right angle. Let ABCD be a circle of which AB is a diameter, and () the centre; draw the chord BD dividing the circle into two segments, viz. BAD greater than, and BCD less than, a semi-circle; join AD, DC, CB. Then ADB in a semi-circle'is a right angle; < BAD “in a segment' greater than a semi-circle is less than a right angle, and - BCD in a segment' less than a semi-circle is greater than a right angle. Join OD; then since OA = OD, LOAD= LODA (26); and since OB=OD, -OBD=- ODB, ..LADB=LOAD + OBD; add to these equals - ADB, then twice 2 ADB a = = =a B = LOAD+ LOBD+ - ADB; .. twice 2ADB=two right angles, (since <OAD, -OBD, and 2 ADB are the three angles of the triangle ABD), and the halves of equal things must be equal, . LADB=one right angle. Also, since LOAD + LOBD= a right angle, ..LOAD, or 2 BAD, which is the same thing, is less than a right angle. Again, by (53), BCD+ ZBAD = two right angles, and since ZBAD is less than a right angle, .in. ZBCD must be greater than a right angle. 55. PROP. VII. A straight line, drawn at right angles to a diameter of a circle from either of its extremities, lies wholly without the circumference of the circle except at that point only. [DEF. A straight line, which lies wholly without the circumference of a circle except at one point only, is said to touch the circle, or to be a tangent to the circle, at that point.] Let a straight line AB be drawn at right angles to AC, a diameter of the circle ACD, from the point A one of its extremities; and let 0 be the centre of the circle. Take E any other point in AB, distinct from A, and join OE; and let OE, or OE produced meet the circumference in F. Then AOE is a triangle; and since the three angles of every triangle are equal to two right angles, and one of them in this case, viz. 20AE is a right angle, . each of the other two angles, OEA, -ĂOE, is less than a right angle; that is, ZOAE is greater than OEA. But the greater side is opposite to the greater angle (33); .. OĚ is greater than 0A, the radius of the circle, that is, OE is greater than OF, or E is without the circumference. And E is any point whatever in AB except A; A B lies wholly without the circumference except at the point A. Cor. 1. The converse of this is also true, viz. that, if a straight line touch the circle at any point, it will be at right angles to the diameter or radius through that D A point. For, if possible, AB being a tangent at A, that is, every point in it except A being without the circle, suppose z OAB not a right angle. From O draw OÉ at right angles to AB, meeting the circumference in F; then OAE is a triangle, of which <OEA is a right angle, and .. 20AE less than a right angle; .. the side 04 is greater than the side OE (33); but OA = OF, .. OF is greater than OE, a part greater than the whole, which is impossible. Hence the supposition that - OAB is not a right angle cannot hold ; : , AB must be at right angles to AO. Cor. 2. Hence, to draw a tangent to a given circle through a given point A in its circumference, find 0 the centre of the circle, join 0A, and draw AB at right angles to OA from the point A ; AB is the tangent required. COR. 3. Hence, also, if a straight line touches a circle, and from the point of contact another straight line be drawn, at right angles to the former, through the circle, the centre of the circle will be in this latter line. 56. PROP. VIII. To draw a tangent to a given circle from a given point without it. Let A be the given point from which it is required to draw a straight line touching the given circle BCD. Find O the centre of the circle (50), and join 0A; E bisect DA in the point E; with centre E and radius EA describe the semi-circle AFO meeting the circle BCD in F; and join AF. AF is the tangent required. For, joining OF, since AFO is a semi-circle, _OFA is a right angle (54), :: AF is at right angles to a radius or diameter of the circle from one of its extremities, .. AF touches the circle at the point F (55). A second tangent may also be drawn from A by |