C E B F D C c Def. 10. 1. equal, then each of the equal angles is a right angle; therefore AFE, BFE, are right angles. Wherefore the right line CD drawn through the centre, bisecting the right line AB not drawn A through the centre, it will also cut it at right angles. But if CD cut AB at right angles, it also bisects it; that is, AF is equal to FB. For, by the same construction, because EA, which is from the centre, is equal to EB, the angle EAF will be equal to the angled EBF, but the right angle AFE is also a 5. 1. equal to the right angle BFE; therefore the two triangles EAF, EBF, have two angles equal, each to each, and one side equal to one side; namely, the side EF common to the two triangles, which is subtended by one of the equal angles. Therefore they will have the remaining sides equal to the remain- © 26. 1. ing sides, and AF will be equal to FB. If, therefore, in a circle a right line drawn through the centre bisect another right line which is not drawn through the centre, it will also cut it at right angles, and if it cut it at right angles, it will also bisect it. Q. E. D. Deduction. If a right line drawn through the centre of a circle bisect any number of right lines which do not pass through the centre, the lines shall be parallel to one another. PROPOSITION IV. If in a circle two right lines cut another, which are not drawn through the centre, they shall not bisect one another. Let ABC be a circle, and in it draw two right lines AC, BD, which cut one another in the point E, and are not drawn through the centre. They do not bisect each other. For if it be possible let them bisect each other, so that AE be equal to EC, and BE to ED, and find the centre of the circle ABCD, which let be A F, and join EF. Therefore because the B right line FE drawn through the centre b F e E a 1. 1. b bisects another right line AC, which is not drawn through bisects another right line BD, which does not pass through the centre, it will also cut it at right angles; therefore FEB is a right angle. But FEA was shown to be a right angle. Wherefore the angle FEA will be equal to the angle FEB, the less to the greater, which is impossible. Therefore AC, BD, do not bisect each other. Wherefore if in a circle two right lines, &c. PROPOSITION V. THEOREM. Q. E. D. If two circles cut one another, they shall not have the same centre. For let the two circles ABC, CDG, cut one another in the point c. They have not the same centre. For if it be possible, let E be the centre, and join Ec, and in the circumference CGD E G F B take any point G, which is not common to both circumferences, EG joined will cut the circumference ACB in F. Because E is the centre of the circle ABC, EC will be equal to EF, and because E is the centre of the circle CDG, EC will be equal to EG; but Ec was shown to be equal to EF: wherefore EF will be equal to EG, the less to the greater, which is impossible. Therefore the point E is not the centre of the circles ABC, CDG. Wherefore if two circles, &c. Q.E.D. PROPOSITION VI. THEOREM. If two circles touch one another internally, they shall not have the same centre. A F E B D For let the two circles ABC, CDE, touch one another internally in the point c. They have not the same centre. For if it be possible, let F be the centre, and join FC, and in the circumference ABC take any point B, which is not common to both circumferences. FB joined shall meet the circumference ECD in E. Therefore because F is the centre of the circle ABC, CF is equal to FB. Again, because F is the centre of the circle CDE, CF will be equal to FE. But CF was shown to be equal to FB; wherefore FE is also equal to FB, the less to the greater, which is impossible. Therefore the point F is not the centre of the circles ABC, CDE. Wherefore if two circles, &c. Q. E. D. PROPOSITION VII. THEOREM. If in the diameter of a circle any point be taken which is not the centre of the circle, and from it any right lines be let fall in the circle, the greatest will be that in which the centre is, and the remainder the least; but of all the others, that which is nearer to that which passes through the centre is greater than that which is more remote, and only two equal right lines from the same point can be drawn in the circle one on each side of the least. B E K H D Let ABCD be a circle whose diameter is AD, and in AD take any point F, which is not the centre of the circle; and let E be the centre of the circle, and from the point F in the circle ABCD draw any right lines FB, FC, FG FA is the greatest, and FD the least; but of the others, FB is greater than FC, and FC greater than FG. For join BE, CE, GE; and because two sides of every triangle are greater than the third; BE, EF, will be greater than BF. But AE is equal to BE; wherefore BE, EF, are equal to AF; AF is therefore greater than FB. Again, because BE is equal to CE, and FE common, the two 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 a 24. 1. than the base FC. For the same reason CF is also greater than FG. Again, because GF, FE, are greater b b 20. 1. than GE, but GE is equal to ED; GF, FE, will be greater than ED, take away the common part FE; therefore the remainder GF is greater than the remainder FD. Therefore FA is the greatest, and FD the least; also BF is greater than FC, and FC than FG. And from the point F only two equal right lines can be let fall into the circle ABCD, one on each side of the least FD. For at the line EF at the given point E in it, make the angle HEF equal to the angle FEG; and join FH. Therefore because GE C is equal to EH, and EF common, the two GE, EF, are equal to the two HE, EF, and the angle GEF is equal a c 23. d 4.1. a 1. 3. to HEF; therefore the base FGd will be equal to the Deduction. If two equal right lines in a circle meet in a point which is not the centre, then the right line which passes through the centre and that point bisects the angle contained by the two equal right lines. PROPOSITION VIII. THEOREM. If without a circle any point be taken, and from it any right lines be drawn to the circle, one of which passes through the centre, and the others any how; of those which fall on the concave circumference the greatest is that which passes through the centre; and of the others, that which is nearer to that which passes through the centre is greater than that which is more remote; and of those which fall on the convex circumference, that is the least which lies between the point and the diameter; and of the others, that which is nearer to the least is less than that which is more remote, and from that point only two equal right lines can be drawn to the circle, one on each side of the least line. Let ABC be a circle, and without it take any point D ; and from it draw to the circle certain right lines DA, DE, DF, DC; and let DA pass through the centre. Of those which fall on the DH. A M e b There- 24. 1. dd Ax. 4. 1. e 21. 1. be м, and join ME, MF, MC, MH, ML, MK. And because AM is equal to ME, and MD is common; wherefore AD is equal to EM, MD; but EM, MD, are greater than ED. b 20. 1. Wherefore AD is also greater than ED. Again, because ME is equal to MF, add мD, which is common, EM, MD, will be equal to MF, MD, but the angle EMD is greater than the angle FMD; therefore the base ED will be greater than the base FD. In like manner we may demonstrate that FD is also greater than CD. fore DA is the greatest, but DE is greater than DF, and DF than DC. Moreover, because мK, KD, are greater than MD, and MK is equal to MG, the remainder KD will be greater than the remainder GD: wherefore GD is less than KD. And because in one side MD of the triangle MLD two right lines are drawn within it, viz. MK, KD, these will be less than ML, LD, of which MK is equal to ML; the remainder, therefore, DK, is less than the remainder DL. In like manner we may show that DL is less than DH. Wherefore DG is the least, but DK less than DL, and DL less than DH. Also only two equal right lines can be drawn from the point D on each side of the least line. Make at the right line MD at the given point м in it, the angle DMB equal to the angle KMD, and join DB. Therefore because мK is equal to £ 23. 1. MB, and MD common, the two Kм, MD, are equal to the two BM, MD, each to each, and the angle KMD is equal to the angle BMD; the base, therefore, DK," is equal to 4. 1. the base DB. From the point D no other right line can fall on the circumference equal to DK. For if it be possible, let DN fall, and because DK is equal to DN, and DK to DB, DB will be also equal to DN; that is, that which is nearer is equal to that which is more remote, which has been proved to be impossible. Therefore if without a circle, &c. Q. E. D. PROPOSITION IX.* THEOREM. If within a circle any point be taken, and from it more than two equal right lines are drawn to the circumference, the point so taken will be the centre of the circle. For within the circle ABC, take any point D, and from the point D let more than two equal right lines DA, DB, * An affirmative demonstration may be, and is, given to this in many editions of the Elements; the present one, although it possesses not that advantage, I deemed preferable both for conciseness and simplicity. |