BOOK III. VI. A segment of a circle is the figure VII. "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 And an angle is said to insist or stand ed between the straight lines that contain the angle. X. The sector of a circle is the figure con- tained by two straight lines drawn XI. Similar segments of a cir cle are those in which PROP. I. PROB. See N. To find the centre of a given circle. a 10. 1. Let ABC be the given circle; it is required to find its centre. Draw within it any straight line AB, and bisecta it in D; 11. 1. from the point D drawb 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 equal to the two BD, DG, each to each; and the base GA is equal to the base GB, because they are drawn from the centre G*: Therefore the angle ADG is equal to the angle GDB: But when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a FG Ꮐ c 8. 1. B d right angled: Therefore the angle GDB is a right angle: 10 Def. 1. But FDB is likewise a right angle: wherefore the angle FDB is equal to the angle GDB, the greater to the less, which is impossible: Therefore G is not the centre of the circle ABC. In the same manner it can be shown, that no other point but F is the centre; that is, F is the centre of the circle ABC: Which was to be found. 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. IF any PROP. II. THEOR. two points be taken in the circumference of a circle, the straight line which joins them shall 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 shall fall within the circle. For, if it do not, let it fall, if possible, without, as AEB; find a 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, D F * 1. 3. b 5. i. E B *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. 16. 1. Book III. a side of the triangle 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 19. 1. is opposited; 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. 1. 3. PROP. III. THEOR. IF a straight line drawn through the centre of a circle bisect a straight line in it which does not pass through the centre, it shall cut it at right angles; and if it cuts it at right angles, it shall 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 cuts it also 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; there8. 1. fore 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, 10 Def. 1. each of them is a right angle: Therefore each of the angles AFE, BFE, is E a right angle; wherefore the straight A F B bisecting another AB that does not pass through the centre, cuts the same at right angles. The same construction being made, because EA, EB, from the centre are equal to one another, the angle EAF 5. 1. is equal to the angle EBF: and the right angle AFE is equal to the right angle BFE: Therefore, in the two tri angles, EAF, EBF, there are two angles in one equal to Book III. two angles in the other, and the side EF, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equale; AF therefore is equal⚫ 26. 1. to FB. Wherefore, if a 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 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. 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, takea F the centre of the circle, and join A EF; and because FE, a straight line through the centre, bisects another AC which does not pass B D a 1. 3 through the centre, it shall cut it at right angles; where. 3. 3. fore FEA is a right angle: Again, because the straight line FE bisects the straight line BD, which does not pass through the centre, it shall 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. PROP. V. THEOR. If two circles cut one another, they shall not have the same centre. Let the two circles ABC, CDG cut one another in the points B, C; they have not the same centre. Book III. For, if it be possible, let E be their centre; join EC, and draw any straight line EFG meet ing them in F and G; and be- G E B Wherefore, if two PROP. VI. THEOR. IF two circles touch one another internally, they shall not have the same centre. Let the two circles ABC, CDE, touch one another inter- E and B; And because F is the E the circles ABC, CDE. Therefore, if two circles, &c. Q. E. D. |