BOOK III. and DE greater than DF, and DF than DC: And because MK, KD are greater than MD, and MK is equal to MG, the remainder KD is greater d than the b 20. 1. d 4. Ax. remainder GD, that is, GD is lefs e 21. 1. H D GB N M than KD And because MK, DK are drawn to the point K within the triangle MLD from M, D, the extremities of its fide MD; MK, KD are less than ML, LD, whereof MK is equal to ML; therefore the remainder DK is lefs than the remainder DL: In like manner it may C be fhewn, that DL is lefs than DH: Therefore DG is the least, and DK lefs than DL, and DL than DH: Also there can be drawn only two equal ftraight lines from the point D to the circumference, one upon each fide of the leaft: At M, in MD, make the angle DMB equal to DMK, and join DB: And because MK is equal to MB, and MD common to the triangles KMD, BMD, the two fides KM, MD are equal to the two BM, MD; and the angle KMD is equal to BMD; theref 4 1. fore the bafe DK is equal to DB: But, befides DB, there can be no ftraight line drawn from D to the circumference equal to DK For, if there can, let it be DN; and because DK is equal to DN, and alfo to DB; therefore DB is equal to DN, that is, the nearer to the leaft equal to the more remote; which is impoffible. If, therefore, any point, &c. Q. E. D. See N. IFA f E PROP. IX. THEOR. A F a point be taken within a circle, from which there fall more than two equal ftraight 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 ftraight 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, there can be drawn only two equal ftraight lines from D to the circumfe rence: rence: but DA, DB, DC are all equal; which is impoffible; Book III. but D is the centre; D therefore is DE G a 7. 3. B Ο NE circumference of a circle cannot cut another D B H If it be poffible, let the circumference ABC cut the circumference DEF in more than two points, viz. in B, G, F; take the centre K of the circle ABC, and join KB, KG, KF And because within the circle DEF there is taken the point K, from which to the circumference DEF fall more than two equal ftraight lines KB, KG, KF, the point K is a the centre of the circle DEF: But K is alfo the centre of the circle ABC; therefore the fame" point is the centre of two circles E that cut one another; which is impoffible. Therefore one circumference of a circle cannot cut another in more than two points. QE. D. IF PROP. XI. THEOR. a 9. 3. b 5. 3. two circles touch each other internally, the straight See N. line which joins their centres, being produced, fhall pass through the point of contact. Let the two circles ABC, ADE, of which ABC is the greater, touch each other internally in the point A, and let F be the centre of the circle ABC, and G the centre of the circle ADE: The ftraight line which joins the centres F, G, being produced beyond G, paffes through the point A. For, if not, let it fall otherwise, if poffible, as FGDH, and join AG and because in CH, the diameter of the circle ABC, there Book III. there is taken the point G which is not the centre, GH is the least line from it to the circumference ; therefore AG is greater a 7. 3. than GH: But AG is equal to GD; therefore GD is greater than GH, the lefs than the greater; which is impoffible. Therefore the ftraight line which joins the points F, G cannot fall otherwise than upon the point A, that is, it must pass through it. Therefore, if two circles, &c. Q. E. D. D G A F E B THEOR. See N. a 8. 3. See N. 2 10.11.1. PROP. XII. F two circles touch each other externally, the ftraight line which joins their centres, fhall pass through the point of contact. Let the two circles ABC, ADE touch each other externally in the point A; and let F be the centre of the circle ABC, and G the centre of ADE: The ftraight line which joins the points F, G fhall pass through the point of contact A. For, if not, let it pafs otherwife, if poffible, as FCDG, and join AG and becaufe G is a point without the circle ABC, GC, which lies between it and a the diameter, is the leaft line B F A D E which is impoffible: Therefore the straight line which joins the points F, G fhall not pass otherwife than through the point of contact A, that is, it muft pafs through it. Therefore, if two circles, &c. Q. E. D. PROP. XIII. THEOR. NE circle cannot touch another in more points than one. ON For, if it be poffible, let the circle EBF touch the circle ABC in more points than one, in the points B, D; join BD, and draw a GH bifecting BD at right angles: Therefore, because the points B, D are in the circumference of each of the circles, circles, the ftraight line BD falls within each of them: And Book. III. ~ their centres are in the straight line GH which bisects BD at b 2. 3. c Cor. 1.3. right angles; therefore GH paffes through the point of contact; d 11. 3. but it does not pass through it, because the points B, D are or 12. 3. without the straight line GH; which is abfurd: Therefore one circle cannot touch another in more points than one. EQUAL PROP. XIV. THEOR. QUAL ftraight lines in a circle are equally ditant from the centre; and thofe which are equally diftant from the centre, are equal to one another. Let the ftraight lines AB, CD, in the circle ABDC, be equal to one another; they are equally diftant from the centre. A b a 1. 3. C 3.3. E D d 47. 1. Take 2 E the centre of the circle ABDC, and from it draw EF, EG perpendiculars to AB, CD: Then, because the straight b12. 1. line EF, paffing through the centre, cuts AB, which does not pafs through the centre, at right angles, it alfo bifects it: Wherefore AF is equal to FB, and AB double of AF. For the fame reason, CD is double of CG: And AB is equal to CD; therefore AF is equal to CG: And because AE is equal to EC, the fquare of AE is equal to the fquare of EC: But the B fquares of AF, FE are equal to the fquare of AE, because AFE is a right angle; and, for the like reafon, the fquares of EG, GC are equal to the fquare of EC: Therefore the fquares of AF, FE are equal to the squares of CG, GE, of which the fquare of AF is equal to the fquare of CG, because AF is equal to CG; therefore the remaining fquare of FE is equal to the remaining square of EG, and the ftraight line FE is therefore equal to EG: But ftraight lines in a circle are faid to be equally diftant from the centre, when the perpendiculars d BOOK III. perpendiculars drawn to them from the centre are equal; Therefore AB, CD are equally distant from the centre. e4. Def. 3. Next, if the ftraight lines AB, CD be equally distant from the centre, that is, if FE be equal to EG; AB is equal to CD; For, the fame conftruction being made, it may, as before, be demonftrated, that AB is double of AF, and CD double of CG, and that the fquares of EF, FA are equal to the fquares of EG, GC, of which the fquare of FE is equal to the fquare of EG, because FE is equal to EG; therefore the remaining fquare of AF is equal to the remaining fquare of CG; and the straight line AF is therefore equal to CG: And AB is double of AF, and CD double of CG; wherefore AB is equal to CD. Therefore equal straight lines, &c. Q. E. D. THE HE diameter is the greateft ftraight line in a circle; and, of all others, that which is nearer to the centre is always greater than one more remote; and the greater is nearer to the centre than the lefs. A B Let ABCD be a circle, of which the diameter is AD, and a 12. J. From the centre draw a EH, EK per- F K H F equal to equal to b b 20. 1. EB, EC: But EB, EC are greater than BC; wherefore, also, AD is greater than BC. d ט D And, because BC is nearer to the centre than FG, EH is 5. Def. 3. lefs than EK: But, as was demonftrated in the preceding, BC is double of BH, and FG double of FK, and the fquares d 47. 1. of EH, HB are equal to the fquares of EK, KF, of which the fquare of EH is lefs than the fquare of EK, because EH is lefs than EK; therefore the fquare of BH is greater than the fquare of FK, and the ftraight line BH greater than FK; and therefore BC is greater than FG. Next, let BC be greater than FG; BC is nearer to the centre than FG, that is, the fame construction being made, EH is less than EK: Because BC is greater than FG, BH likewise |
