"The angle of a segment is that which is contained by the "straight line and the circumference." VIII. An angle in a segment is the angle con- IX. And an angle is said to insist or stand X. The sector of a circle is the figure contain- See N. a 10. 1. b 11. 1. XI. Similar segments of a circle PROP. I. PROB. To find the centre of a given circle. Let ABC be the given circle; it is required to find its centre. b Draw within it any straight line AB, and bisect a it in D; from the point D draw DC at right angles to AB, and produce it to E, and bisect CE in F: The point F is the centre of the circle ABC. For, if it be not, let, if possible, G be the centre, and join Book III. GA, GD, GB: Then because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG are A F. C G c 8. 1. B ID E equal to the two BD, DG, each to COR. From this it is manifest, that if in a circle a straight line bisect another at right angles, the centre of the circle is in the line which bisects the other. PROP. II. THEOR. If any two points be taken in the circumference of a circle, the straight line which joins them must fall within the circle. Let ABC be a circle, and A, B any two points in the circumference; the straight line drawn from A to B must fall within the circle. For, if it do not, let it fall, if possible, without, as AEB; find D the centre of the circle ABC, and join AD, DB, and produce DF, any straight line meeting the circumference AB, to E: Then because DA is equal to DB, the angle DAB is equal to the angle DBA; and because AE, a side of the triangle D C d 10. Def. 1. a 1. 3. F b 5. 1. N. B. Whenever the expression "straight lines from the centre," or "drawn from the centre," occurs, it is to be understood that they are drawn to the circumference. c 16. 1. d 19. l. Book III. DAE, is produced to B, the angle DEB is greater than the angle DAE; but DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE: but to the greater angle the greater side is opposite; DB is therefore greater than DE: But DB is equal to DF; wherefore DF is greater than DE, the less than the greater, which is impossible: Therefore the straight line drawn from A to B does not fall without the circle. In the same manner, it may be demonstrated, that it does not fall upon the circumference: it falls therefore within it. Wherefore, "if any two points," &c. Q. E. D. a 1. 3. b 8. 1. PROP. III. THEOR. If a straight line drawn through the centre of a circle bisect a straight line in the circle, which does not pass through the centre, it will cut that line at right angles; and if it cut it at right angles it will bisect it. Let ABC be a circle; and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre, in the point F: It also cuts it at right angles. Take a E the centre of the circle, and join EA, EB. Then, because AF is equal to FB, and FE common to the two triangles AFE, BFE, there are two sides in the one equal to two sides in the other, and the base EA is equal to the base EB; therefore the angle AFE is equal to the angle BFE. But when a straight line standing upon another, makes the adjacent angles equal to one another, each of them is c 10. Def. 1. a right angle: Therefore each of the angles AFE, BFE is a right angle; wherefore the straight line CD, drawn through the centre bisecting another AB that does not pass through the centre, cuts the same at right angles. d 5. 1. E F B D But let CD cut AB at right angles; The same construction being made, because EA, EB from there are two angles in the one equal to two angles in the other, Book III. and the side EF, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equal; AF therefore is equal to FB. Wherefore, "if a e 26. 1. "straight line," &c. Q. E. D. PROP. IV. THEOR. If in a circle two straight lines cut one another which do not both pass through the centre, they do not bisect each the other. Let ABCD be a circle, and AC, BD two straight lines in it, which cut one another in the point E, and do not both pass through the centre: AC, BD do not bisect one another. A F D a 1. 3. For, if it is possible, let AE be equal to EC and BE to ED: If one of the lines pass through the centre, it is plain that it cannot be bisected by the other which does not pass through the centre: But if neither of them pass through the centre, take a F the centre of the circle, and join EF: and because FE, a straight line through the centre, bisects another AC, which does not pass through the centre, it must cut it at right b angles; wherefore FEA is a right angle. Again, because the straight line FE bisects the straight line BD, which does not pass through the centre, it must cut it at right angles; wherefore FEB is a right angle: and FEA was shown to be a right angle; therefore FEA is equal to the angle FEB, the less to the greater, which is impossible: Therefore AC, BD do not bisect one another. Wherefore, "if in a circle," &c. Q. E. D. B E PROP. V. THEOR. If two circles cut one another, they cannot have the same centre. Let the two circles ABC, CDG cut one another in the points B, C; they have not the same centre. b 3. 3. Book III. A For, if it be possible, let E be their centre: Join EC, and draw any straight line EFG meeting the circles in F and G; and because E is the centre of the circle ABC, CE is equal to EF: Again, because E is the centre of the circle CDG, CE is equal to EG: But CE was shewn to be equal to EF; therefore EF is equal to EG, the less to the greater, which is impossible: Therefore E is not the centre of the circles ABC, C D Ꮐ F E B CDG. Wherefore, "if two circles," &c. Q. E. D. PROP. VI. THEOR. If two circles touch one another internally, they cannot have the same centre. Let the two circles ABC, CDE, touch one another internally in the point C: They have not the same centre. For if they have, let it be F; join FC, and draw any straight line FEB meeting the circles in E and B; and because F is the F C B E D |