Book III. VI. A fegment of a circle is the figure con- VII. "The angle of a fegment is that which is contained by the "ftraight line and the circumference." VIII. Let ABC be the given circle; it is required to find its center. Draw within it any ftraight line AB, and bisect a it in D; from the point D drawb DC at right angles to AB, and produce it to E, and bifect CE in F: The point F is the center of the circle ABC. For For, if it be not, let, if poffible, G be the center, and join Book III. GA, GD, GB: Then, because DA is equal to DB, and DG C FG C 8. I D B E common to the two triangles ADG, COR. From this it is manifeft, that if in a circle a straight line bifect another at right angles, the center of the circle is in the line which bifects the other. PROP. II. THEOR. F any two points be taken in the circumference of a within the circle. Let ABC be a circle, and A, B any two points in the ciṛcumference; the ftraight line drawn from A to B fhall fall within the circle. For, if it do not, let it fall, if poffible, without, as AEB; find D the center of the circle ABC, and join AD, DB, and produce DF any ftraight 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 fide of the triangle E 3 C d Io. def. I a 1. 3. F bs. I. A A E B DAE, +N. B. Whenever the expreffion " ftraight lines from the center," or " drawn "from the center," occurs, it is to be understood that they are drawn to the circumference. C 16. I. dr. I. 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 fide is oppofite; 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 impoffible: Therefore the straight line drawn from A to B does not fall without the circle. In the fame 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 I. 3. b 8. I. IF PROP. III. THE OR. F a ftraight line drawn through the center of a circle bifect a ftraight line in it which does not pass through the center, it fhall cut it at right angles; And if it cuts it at right angles, it fhall bifect it. Let ABC be a circle; and let CD, a ftraight line drawn through the center bifect any straight line AB, which does not pafs through the center, in the point F: It cuts it also at right angles. Take 2 E the center 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 fides in the one equal to two fides in the other, and the base EA is equal to the bafe EB; therefore the angle AFE is equal to the angle BFE: But when a ftraight line standing upon another makes the adjacent angles equal to one another, each of them is a right c 10. def. 1. angle: Therefore each of the angles AFE, BFE is a right angle; wherefore the ftraight line CD, drawn through the center bifecting another AB that does A not pass thro' the center, cuts the fame at right angles. d 5.1. E Ꭰ B But let CD cut AB at right angles; CD alfo bifects it, that is, AF is equal to FB. The fame conftruction being made, becaufe EA, EB from the center are equal to one another, the angle EAF is equal to the angle EBF; and the right angle AFE is equal to the right angle BFE: Therefore, in the two triangles EAF, EBF, there there are two angles in one equal to two angles in the other, Book III. and the fide EF, which is opposite to one of the equal angles in in each, is common to both; therefore the other fides are e qual; AF therefore is equal to FB. Wherefore, if a straight e 26. 1. line, &c. Q. E. D. PROP. IV. THEOR. Thin a circle two straight lines cut one another which do not both pass through the center, they do not bifect 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 center: AC, BD do not bifect one another. a For, if it is poffible, let AE be equal to EC, and BE to ED: If one of the lines pafs thro' the center, it is plain that it cannot be bifected by the other which does not pass thro' the center: But, if neither of them pafs thro' the center, take F the center of the circle, and join EF: And because FE, a ftraight line thro' the center, bifects another A AC which does not pafs thro' the center, it fhall cut it at right angles; B wherefore FEA is a right angle: Again, becaufe the ftraight line FE bi F D a I. 3. E C b 3.3. fects the ftraight line BD which does not pass thro' the center, it shall cut it at right angles; wherefore FEB is a right angle: And FEA was fhewn to be a right angle; therefore FEA is equal to the angle FEB, the lefs to the greater, which is impoffible: Therefore AC, BD do not bifect 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 fame center. Let the two circles ABC, CDG cut one another in the points B, C; they have not the fame center. E 4 For, Book III. C G D E For, if it be poffible, let E be their center: Join EC, and draw any ftraight line EFG meeting them in F and G: And because E is the centre of the circle ABC, CE is equal to EF: Again, becaufe E is the center of the circle A CDG, CE is equal to EG: But CE was fhewn to be equal to EF; therefore EF is equal to EG, the lefs to the greater, which is impoffible: Therefore E is not the center of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D. IF PROP. VI. THE OR... F two circles touch one another internally, they shall not have the fame center. Let the two circles ABC, CDE touch one another internally in the point C: They have not the fame center. For, if they can, let it be F; join FC and draw any straight line FEB meeting them in E and B: And because F is the center of the F E D B PROP. |