reach the circumference; therefore the right line AB is within Book III. the circle. Wherefore, &c. COR. Hence, if a right line touches a circle, it will touch it only in one point. PRO P. III. THEO R. IF, in another line not drawn through the centre, into two equal parts, it ball cut it at right angles; and if it cut it at right angles, it fball cut it into two equal parts. Let ABC be the given circle, CD the line paffing through the centre, cutting the right line AB, not paffing through the centre, into two equal parts, it will cut it at right angles; and if the angles AFE, EFB, be right angles, AF will be equal to FB. c 8. I. I. For, find the centre Ea; join EA, EB; for, because AF is e- a 1¿ qual to FB, and FE is common, the two fides AF, FE, are equal to BF, FE, and the bafe AE equal to EBb; therefore, the angle b def. 15.1. AFE is equal to EFB, and each is a right angled; therefore, d Def. 1o. CD cuts the right line AB at right angles. If EFA, EFB, are right angles, AF is equal to FB; for, because AE is equal to EB, and EF common, the two fides AE, EF, are equal to BE, e Hyp. EF, and AFE, EFB, right angles, and the angle EAB equal to f 5. 1. EBA; therefore, the remaining angle AEF is equal to FEB; g Cor. za therefore the base AF is equal to FB". Wherefore, &c. PROP. IV. THE OR. E two right lines are drawn in a circle, neither of them paffing through the centre, they will not mutually bifect each other. Let two right lines AC, BD, not paffing through the center, be drawn in the circle ABCD, cutting each other in the point E, they will not mutually bifect each other; for, if poffible, let AE be equal to EC, and BE to ED; find the center F, and join FE; then, becaufe FE, paffing thro' the center, cuts the right line AC, not paffing through the center, into equal parts, the angles FEA, FEC, are right angles; and, because FE bifects BD, FEB is a right angle"; therefore, FEA is equal to FEB, a part to the whole, which is impoffible; therefore, AC, BD, do not mutually bifect each other. Wherefore, &c. I. h 4. I. a 3 PROP. BOOK III. a Def. 15. PRO P. V. THE OR. IF two circles cut each other, they cannot bave the same center. Let the two circles ABC, CDG, cut each other, they cannot have the fame center. If poffible, let E be the center of both, draw CE to the point of fection C, and EFG through any other point; then, becaufe E is the center of the circle ABC, EC is equal to EF; and, becaufe E is the center of the circle CDG, EC is equal to EG; b Ax. 11. therefore, EF is equal to EG, a part to the whole, which is impoffible; wherefore E is not the center of both. Wherefore, &c. I. IF Let the two circles ABC, CDE, touch each other inwardly in the point C, they have not the fame center. If poffible, let it be F; join FC, and draw FB through any other point. Then, because F is the center of the circle ABC, CF is equal a def. 15.1. to FB ; for the fame reafon CF is equal to FE; therefore FE is equal to FB, that is, a part to the whole, which is impoffible. Wherefore, &c. b ax. I. IF fome point is taken in the diameter of a circle, which is not the center, from that point if several right lines are drawn te the circumference, the greatest of these right lines is that part of the diameter in which the center is, and the remainder of the diameter is the least; of the other lines, the nearest to that passing through the center is greater than that more remote; and, on each fide of the diameter, only two right lines can be drawn from that point to the circumference equal to one another. Let F be a point in the diameter of the circle ABCD, which is not the center, and from it be drawn FA thro' the center, and FB, FC, FG, any how to the circumference, FA is the greateft line, FA is greater than FB, FB greater than FC, FC great ་་ er than FG, and FD the leaft. And from the point F only two Book III. right lines can be drawn equal to one another on each fide of diameter for, find the center E; join BE, CE, and GE. Then, becaufe E is the center of the circle, EA is equal to EB; add EF to both; then AF is equal to BE, EF; but BE, EF, are greater than BF2; therefore AF is greater than BF; but BE is a 20. 1. equal to CE, and EF common; therefore BE, EF, are equal to CE, EF; but the angle BEF is greater than CEFb; therefore, the b Ax, 9. bafe BF is greater than CF. For the fame reafon CF is greater c 24. 1. than GF; likewife the two fides GF, FE, of the triangle GEF, are greater than GE, that is, than ED; take EF from both, there remains GF greater than FDd; therefore, AF is the great- d Ax. 5. 1. eft right line, and FD the leaft. Laftly, on each fide of the diameter, from the point F, only two right lines can be drawn equal to one another; for, at the point E, with the right line EF, make the angle FEH equal to FEG, then the bafe FH is equal to GF. If any other right e 4. I. line can be equal to FG, let it be FK, that is, a line nearer to that paffing through the center, equal to one more remote, which cannot be. Wherefore, &c. PRO P. VIII. THE OR. IF a point be taken without the circle, and from it right lines be drawn, one of which passing through the center, and the other falling upon the concave part of the circumference, the greatest of thefe lines is that paffing through the center; and the line nearer to that, paffing through the center, is greater than that more remote; of thofe falling upon the convex part of the circumference, that which lies betwixt the point and the diameter is the least line, and that line nearer to that paffing through the center, is less than that more remote; and, on each fide of the diameter, only two lines can be drawn from that point, falling either on the concave or convex part of the circumference equal to one another. Let any point D be taken without the circle ABC; draw DA, DE, DF, DC, to the concave part of the circumference; of these lines DA, which paffes through the center, is the greateft; DE is greater than DF, and DF than DC. Of these that fall upon the convex part of the circumference, DG is the leaft, DK is lefs than DL, and DL than DH; on each fide of DG only two right lines can be drawn equal to each other, either on the convex or concave part of the circumference. For, find the center M; draw ME, MF, MC, MH, ML, MK. Now, becaufe MA is a 20. 1. b 24. I. Book III. is equal to ME, add MD, which is common to both, then AD is equal to DM, ME; but DM, ME, are greater than DE; therefore DA is greater than DE; but DM, ME, are equal to DM; MF, and the angle DME greater than DMF; therefore DE is greater than DF. For the fame reafon, DF is greater than DC; wherefore DA is the greatest of the right lines falling on the concave part of the circumference. Again, because DK c Def. 15. KM are greater than DM, take the equal lines KM, GMC, from both, there remains DG less than DK; but K is a point taken within the triangle DLM; therefore DK, KM, are lefs than DL, LM; take MK, ML from both, there remains DK lefs than DL. For the fame reafon, DL is lefs than DH; wherefore DG is the leaft line, and DK lefs than DL, &c. I. d Ax. 5. I. € 21. I. € 13. 4. Likewise, from the point D on each fide of the leaft line, only two right lines can be equal to each other, falling on the convex part of the circumference. For, make the angle DMB equal to the angle DMK, and join DB; then, because DM, MB, are equal to DM, MK, and the angle DMB to DMK, the base DB is equal to DK. If any other right line can be equal to DB, let it be DN; that, is a line nearer to the least line equal to one more remote, which cannot be. Neither can more than two equal right lines fall upon the concave part of the circumference on each fide of the diameter from the fame point. For, let the angle AMO be made equal to AME, join MO, DO; then the angles AMO, DMO, are equal to two right angles, and AME, DME, likewife equal to two right angles; but AMO h Ax. 3. 1. is equal to AME; therefore, the angle DMO is equal to DME"; but DM, ME, are equal to DM, MO, and the angle DME to DMO; therefore, the bafe DO is equal to DE. If any other right line can be equal to DO, let it be DP, that is, one nearer to that paffing through the center, equal to one more remote. Wherefore, &c. i 4. z. PRO P. IX. THE OR. TF a point be affumed in a circle, and from it be drawn more than two right lines to the circumference equal to one another, that point is the center of the circle. Let the point D be affumed in the circle ABC, and from it be drawn, to the circumference, the right lines DB, DC, DA, equal to one another, D is the center of the circle. If not, let E be the center, join D, E, which produce to F and G; then is FG the diameter, and D is fome point in it, not the center therefore DG is greater than DC, DC than DB, and DB Book III.. than DA'; but DC, DB, DA, are equal; and likewife not equal; which is impoffible; therefore no point but D is the cen- a 7. ter of the circle. Wherefore, &c. PRO P. X. THE OR. b Hyp. O NE circle cannot cut another in more than two points. For, if poffible, let the circle ABC cut the circle DEF in the points B, G, F; let K be the center of the circle ABC; join BK, KG, KF. Now, because K is a point within the circle DEF, from which there is drawn to the circumference the right lines BK, KG, KF, equal to one another; therefore K is the center of both circles; which is impoffible b. Wherefore, a 2. &c. I PRO P. XI. THE OR. F two circles touch one another inwardly, a line joining their centers will fall on the point of contact. b 5. Let the two circles ABC, ADE, touch each other inwardly er than AF b; therefore DF is greater than AF; therefore 20. 1. I PRO P. XII. THE OR. F two circles touch one another outwardly, a line joining their centers will pass through the point of contact. Let the two circles ABC, ADE, touch one another outwardly in the point A; a right line, joining their centers, will |