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. For, if it is possible, let af 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, D take (111. 1.) F the centre of the circle, and join EF: and because F E, a straight A line through the centre, bisects another AC which does not pass through the centre, it shall cut it at right (111. 3.) 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 shall cut it at right (m. 3.) 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 В 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. For, if it be possible, let e be their centre: Join EC, and draw any straight line EFG meeting them in F and G: And because E is the centre of the circle ABC, CE is equal to EF: Again, because E is the G centre of the circle CDG, CE is equal A A to EG: But CE was shown to be equal to EF, therefore EF is equal to EG, the less to the greater, which is impossible: Therefore E is not the E centre of the circles ABC, CDG. Wherefore if two circles, &c. Q.E.D. 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 internally in the point c: They have not the same centre. 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 centre of the circle ABC, CF is equal to FB: Also because F is the centre of the circle B CDE, CF is equal to FE: And CF was А shown equal to FB; therefore FE is equal to FB, the less to the greater, which is impossible; wherefore F is D PROP. VII. THEOR. If any point be taken in the diameter of a circle which is not the centre, of all the straight lines which can be drawn from it to the circumference, the greatest is that in which the centre is, and the other part of that diameter is the least; and, of any others, that which is nearer to the line which passes through the centre is always greater than one more remote : and from the same point there can be drawn only two straight lines that are equal to one another, one upon each side of the shortest line. Let ABCD be a circle, and Ad its diameter, in which let any point F be taken which is not the centre: Let the centre be E; of all the straight lines FB, FC, FG, &c., that can be drawn from F to the circumference, FA is the greatest, and FD, the other part of the diameter Ad, is the least: and of the others, FB is greater than FC, and FC than FG. Join BE, CE, GE; and because two sides of a triangle are B H D greater (1. 20.) than the third, BE, EF are greater than BF; but AE is equal to EB; therefore AE, EF, A that is AF, is greater than by: Again, because BE is equal to CE, and FE common to the triangles BEF, CEF, the two sides BE, EF are equal to the two CE, EF; but the angle BEF is greater than the angle CEF; therefore the base BF is greater (1. 24.) than the base FC: For K the same reason, CF is greater than GF: Again, because GF, FE are greater (1. 20.) than Eg, and Eg is equal to ED; therefore GF, FE are greater than Ed: Take away the common part FE, and the remainder GF is greater than the remainder FD: Therefore fa is the greatest, and FD the least of all the straight lines from F to the circumference; and BF is greater than CF, and CF than GF. Also there can be drawn only two equal straight lines from the point F to the circumference, one upon each side of the shortest line FD: At the point E in the straight line E F, make (I. 23.) the angle Fer equal to the angle GEF, and join FH: Then because GE is equal to EH, and EF common to the two triangles G EF, HEF; the two sides GE, EF are equal to the two HE, EF; and the angle GEF is equal to the angle HEF; therefore the base Fg is equal (1. 4.) to the base Fh: But, besides Fi, no other straight line can be drawn from F to the circumference equal to FG: For, if there can, let it be FK; and because 'FK is equal to FG, and Fg to FH, FK is equal to Fh; that is, a line nearer to that which passes through the centre, is equal to that which is more remote ; which is impossible. Therefore, if any point be taken, &c. Q.E.D. PROP. VIII. THEOR. If any point be taken without a circle, and straight lines be drawn from it to the circumference, whereof one passes through the centre; of those which fall upon the concave circumference, the greatest is that which passes through the centre; and of the rest, that which is nearer to the one passing through the centre is always greater than one more remote: But of those which fall upon the convex circumference, the least is that between the point without the circle ; and the diameter ; and of the rest, that which is nearer to the least is always less than one more remote : And only two equal straight lines can be drawn from the same point to the circumference, one upon each side of the least line. Let ABC be a circle, and D any point without it, from which let the straight lines DA, DE, DF, DC be drawn to the circumference, whereof DA passes through the centre. Of those which fall upon the concave part of the circumference A EFC, the greatest is AD which passes through the centre; and the nearer to it is always greater than the more remote. viz. DE than DF, and dr than Dc: But of those which fall upon the convex circumference HLKG, the least is DG betwen the point D and the diameter ag; and the nearer to it is always less than the more remote, viz. DK than DL, and DL than DH. Take (II. 1.) m the centre of the circle ABC, and join ME, MF, MC, MK, ML, MH: And because Am is equal to ME, add MD to each, therefore ad is equal to EM, MD; but EM, MD are greater (1. 20.) than ED; therefore also ad is greater than ED: Again, because me is equal to MF, and md common to the triangles EMD, FMD; EM, MD D are equal to FM, MD; but the angle EMD is greater than the angle FMD; therefore the base Ep is greater (1. 24.) than the base FD: In like manner it may be shown that rd is greater than cD: Therefore DA is the greatest; and DE greater than DF, and DF than DC: And because MK, KD are greater than MD, and C MK is equal to MG, the remainder KD is greater (Ax. 5.) than the remainder GD, that is, hd is less than KD: And because MK, DK are drawn to the point K within the triangle MLD, from M, D, the extremities of its side MD, therefore MK, KD are less (1.21.) than ML, LD, whereof MK is equal to ML; therefore the remainder Dk is less than the remainder Dl: In like manner it may be shown, that dl is less than th: Therefore Dg is the least, and dk less than DL, and DL than DA: Also there can be drawn only two equal straight lines from the point d to the circumference, G B H А one upon each side of the least : At the point m, in the straight line md, make (1. 23.) the angle DMB equal to the angle DMK, and join DB: And because m K is equal to MB, and MD common to the triangles KMD, BMD, the two sides KM, MD are equal to the two BM, MD; and the angle KMD is equal to the angle BMD; therefore the base dk is equal (1. 4.) to the base dB: But, besides DB, there can be no straight line drawn from to the circumference equal to DK: For, if there can, let it be Dn; and because Dk is equal to dn, and also to DB; therefore DB is equal to Dn, that is, the nearer to the least equal to the more remote, which is impossible. If, therefore, any point, &c. Q. E. D. PROP. IX. PROB. If a point be taken within a circle, from which there fall more than two equal straight lines to the circumference, that point is the centre of the circle. Let the point d be taken within the circle ABC, from which to the circumference there fall more than two equal straight lines, viz. DA, DB, DC, the point D is the centre of the circle. For, if not, let E be the centre, join De and produce it to the circumference in F, G; then Fg is a diameter of the circle ABC: And because in FG, the diameter of the circle ABC, there is taken the point D which is not the centre, DG shall be the great F G est line from it to the circumference, and pc greater (11. 7.) than DB, and DB than DA: But they are likewise equal B (Hyp.), which is impossible: Therefore È is not the centre of the circle ABC: In like manner it may be demonstrated that no other point but d is the centre; D therefore is the centre. Wherefore, if a point be taken, &c. Q. E. D. PROP. X. THEOR. One circumference of a circle cannot cut another in more than two points. If it be possible, let the circumference FAB cut the cir |