$ 19. 1. * 1. 3. 8. 1. 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. PROPOSITION 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 cut it at right angles, it shall besect 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* 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 But when a straight line standing upon another makes the adjacent angles equal to one another, each of 10 Def. 1. them is a right angle; therefore E B 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. But let CD cut AB at right angles: CD also bisects it, that is, AF is equal to FB. The same construction being made, because EA, EB || 15 Def. 1. 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 triangles EAF, EBF, there are two angles in one equal to 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 equal¶; AF therefore is equal to FB. ¶ 26. 1. Wherefore, if a straight line, &c. Q. E. D. PROPOSITION 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, take* F the centre of the circle, and join EF: and because FE, a straight line through the centre, bisects another AC which * 1. 3. does not pass through the centre, it shall cut it at right† † 3. 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 angles; wherefore FEB is a right angle: 3. 3. And FEA was shown to be a right angle: therefore FEA is equal to the angle FEB, the less to the greater, which || 1 AX. is impossible: therefore AC, BD do not bisect one another. Wherefore, if in a circle, &c. Q. E. D. PROPOSITION 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 A any straight line EFG meeting C them in F and G: and because E is the centre of the circle ABC, Again, because E is the centre of the circle CDG, CE +15 Def. 1: is equal 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 centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D. PROPOSITION 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 have, 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 * 15 Def. 1. the circle ABC, CF is equal* to FB; also, because F is the centre +15 Def. 1. of the circle CDE, CF is equal† to FE: I 1 Ax. And CF was shown to be equal D B to FB; therefore FE is equal to FB, the less to the greater, which is impossible: wherefore F is not the centre of the circles ABC, CDE. Therefore, if two circles, &c. Q. E. D. PROPOSITION 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 greater* than * 20. 1. the third, BE, EF are greater than BF: But AE is equal† to EB; therefore AE, EF, that is AF, is greater than BF: B L + 15 Def. 1. 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 than the | 24. 1. base FC: K G D. For the same reason, CF is greater than GF: 19 Ax. Again, because GF, FE are greater§ than EG, and EG § 20. 1. is equal to ED; GF, FE are greater than ED: take away the common part FE, and the remainder GF is greater ¶¶3 Ax. 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 EF, make* the angle FEH equal to the angle GEF, and • 23. 1. join FH: Then because GE is equal to EH, and EF common to the two triangles GEF, HEF, the two sides GE, EF are equal to the two HE, EF; and the angle GEF is equal† † Constr. to the angle HEF; therefore the base FG is equal to the ‡ 4. 1. base FH: But, besides FH, 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 || 1. Ax which passes through the centre, is equal to one which is more remote; which is impossible. Therefore, if any point be taken, &c. Q. E. D. PROPOSITION 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 con G ག 20. 1. cave circumference, the greatest is that which passes through the centre; and of the rest, that which is nearer to that through the centre is always greater than the 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 the more remote: and only two equal straight lines can be drawn from the point to the circumference, one upon each side of the least. 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 AEFC, 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 DF than DC: but of those which fall upon the convex circumference HLKG, the least is DG, between 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* 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 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 are equal to FM, MD: but the angle EMD is greater than the angle FMD; therefore the base ED is greater than the base FD. In like manner it may be shown that FD is greater than CD: there- F fore DA is the greatest; and DE greater than DF, and DF than DC. And because MK, KD are greater H E ט. GB N M than MD, and MK * 15 Def. 1. is equal* to MG, the remainder KD is greater† than the remainder GD; that is, GD is less than KD: + 5 Ax. 21. 1. And because MK, DK are drawn to the point K within the triangle MLD, from M, D, the extremities of its side MD, MK, KD are less than ML, LD, whereof MK is |