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. IF, in in a circle, a right line be drawn through the centre, cutting another line not drawn through the centre, into two equal parts, it sball cut it at right angles ; and if it cut it at right angles, it shall cut it into two equal parts. Let ABC be the given circle, CD the line passing through the centre, cutting the right line AB, not passing 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. For, find the centre E“; join EA, EB; for, because AF is e. a 1. qual to FB, and FE is common, the two sides AF, FE, are equal to BF, FE, and the base AE equal to EBb; therefore, the angle b def. 15. s. AFE is equal to EFB', and each is a right angled, therefore, Def. 10: 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 sides AE, EF, are equal to BE, e Hyp. EF, and AFE, EFB, right anglese, and the angle EAB equal to f 5.1. EBAf ; therefore, the remaining angle AEF is equal to FEB8 ; & Cor. 3x therefore the base AF is equal to FB". Wherefore, &c. h 4. I PRO P. IV. THEO R. IF, through the centre, they will not mutually bisect each other. Let two right lines AC, BD, not passing through the center, be drawn in the circle ABCD, cutting each other in the point E, they will 1100 mutually bisect each other ; for, if possible, let AE be equal to EC, and 'BE to ED; find the center F, and join FE; then, because FE, passing thro' the center, cuts the right line AC, not passing through the center, into equal parts, the angles FEA, FEC, are right anglesa ; and, because FE bisects BD, * 3: FEB is a right angle; therefore, FEA is equal to FEB, a part to the whole, which is impoflible; therefore, AC, BD, do not mutually bifect each other. Wherefore, &c. PROP. Book III. PRO P. V. THE O R. IF two circles cut each other, they cannot bave the sanie center. Let the two circles ABC, CDG, cut each other, they cannot have the same center. If possible, let E be the center of both, draw CE to the point of sextion C, and EFG through any other point; then, becaufe a Def. 15. E is the center of the circle ABC, EC is equal to EFa ; and, be., cause E is the center of the circle CDG, EC is equal to EG; b Ax. II. 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 F two circles touch each other inwardly, they have not the Jame center. Let the two circles ABC, CDE, touch each other inwardly in the point C, they have not the same center. If possible, 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 same reason CF is equal to FE; therefore FE' is equal to FBb, that is, a part to the whole, which is impossible. Wherefore, &c. bax. 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 to 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 pasing through the center is greater than that more remote ; and, on each side 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 least. And from the point F only two Book III. right lines can be drawn equal to one another on each side of diameters for, find the center E ; join BE, CE, and GE. That, because 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 BF*; therefore AF is greater than BF; but BE is a 20. s. 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. basé BF is greater than CF¢. For the same reason CF is greater c 24. 8. than GF; likewise the two sides 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 least. Lastly, on each side 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 base FH is equal to GFe. If any other right ¢ 4. I. line can be equal to FG, let it be FK, that is, a line nearer to that passing through the center, equal to one mote remote, which cannot be. Wherefore, &c. PRO P. VIII. THEOR. IF a point be taken without the circle, and from it right lines be drawn, one of which paling through the center, and the other." falling upon the concave part of the circumference, the greatest of these lines is that pasing through the center ; and the line neater to that, passing through the center, is greater than that more remote ; of those 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 passing through the center, is less than that more remote ; and, on each side 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 pcine D be taken without the circle ABC; draw DA, DE, DF, DC, to the concave part of the circumference; of these lines DA, which passes through the center, is the greatest; DE is greater than DF, and DF than DC. Of these that fall upon the convex part of the circumference, DG is the least, DK is less than DL, and DL than DH; on each side 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, because MA is a 20. 1. € 21. 1. 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 b 24. I, greater than DFb. For the same reason, 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 d Ax. s. I. both, there remains DG less than DKd; but K is a point taken within the triangle DLM; therefore DK, KM, are less than DL, LM; take MK, MLC from both, there remains DK less than DL, For the same reason, DL is less than DH; wherefore DG is the least line, and DK less than DL, &c. Likewise, from the point D on each side of the least 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 bafe DB 14. 30 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 side of the diameter from the fame point. For, let the angle AMO be made equal to AME, join MO, $ 13. DO; then the angles AMO, DMO, are equal to two right angles, and ÁME, DME, likewise equal to two right angies, but AMO h Ax. 3. I. is equal to AME; therefore, the angle DMO is equal to DMES; but DM, ME, are equal to DM, MO, and the angle DME to i DMO; therefore, the base DO is equal to DEI. If any other right line can be equal to DO, let it be DP, that is, one nearer to that passing through the center, equal to one more remote, Wherefore, &c. PRO P. IX. THEO R. IF a point be assumed 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 assumed 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 Fand G; then is FG the diameter, and D is some point in it, not the center 3 therefore DG is greater than DC, DC than DB, and DB Book III. . than DA'; but DC, DB, DA, are equal", and likewise not equal; which is impossible; therefore no point but D is the cen-a. ter of the circle. Wherefore, &c. ъНур. PRO P. X. THE O R. ONE NE circle cannot cut another in more than two points. For, if poflible, 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 circlesa; which is impossibleb. Wherefore, &c. PRO P. XI. THEO R. I Ftwo circles touch one another inwardly, a line joining their centers will fall on the point of contact. Let the two circles ABC, ADE, touch each other inwardly in the point A ; let F and G be the centers of the circles ABC, ADE; then the line joining the centers F, G, will pass through the point A. If not, let the right line joining the centers F, G, cut the circles in the points D, H; join GA; then because F is the center of the circle ABC, FA is equal to FHa. For the a def. 15. 1. same reason, GD is equal to GA; but GA, GF, are greater than AFb; therefore DF is greater than AF; therefore b 20.1. greater than HF; and likewise less “; which is impossible;c Ax. 9. 1. therefore the line joining the centers will not pass through any other point than A. Wherefore, &c. PRO P. XII. THEO R. Ftwo circles touch one another outwardly, a line joining I . Let the two circles ABC, ADE, touch one another outWardly in the point A ; a right line, joining their centers, will |