Book III. PROP. 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 passing 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 a than the third, BE, EF are greater than BF; but a 20. 1. AEis equal to EB; therefore AE, EF, that is AF, is greater than BF: 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 greaterb b 21. 1. than the base FC: For the same reason, CF is greater than GF; Again, because GF, FE are greater a than EG, and EG is equal to ED; GF, FE are greater than ED: Take away the common part FE, and the remainder GF is greater 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 H Book III. shortest line FD: At the point E in the straight line EF, S o makec the angle FEH equal to the angle GEF and join c 23. 1. 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 to the d 4. 1. angle HEF; therefore the base BG is equal to the base FH: But, besides FH, no 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 which passes through the centre, is equal to one more remote, which is impossible. Therefore, " if any point be taken," &c. Q. E. D. because he equal to rest line can PROP. 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 concave 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 unto 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 line nearer to AD 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 a M the centre of the circle ABC, and join ME, MF, Book III. MC, MH, ML, MK: And because AM is equal to ME, add MD to each, therefore AD is equal to EM, MD: but EM, MD a 1. 3. are greater than ED; therefore also AD is greater than ED: b 20. 1. 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 C than the c 24. 1. base FD: In like manner, it may be shown, that FD is greater than CD: Therefore DA is the greatest; and DE greater than DF, and DF than DC: And because MK, KD, are greater b than MD, and MK is equal to MG, the remainder KD is greater ac d 5. Ax. than the remainder DG, that is, GD is less than KD: And because MK, DK are drawn to the 'FU point K within the triangle MLD from M, D, the extremities of its side MD, MK, KD, are less than ML, LD, whereof e 21. 1. MK is equal to ML; therefore the remainder DK is less than the remainder DL: In like manner, it may be shown, that DL is less than DH: Therefore DG is the least, and DK less than DL, and DL than DH: Also, there can be drawn only two equal straight lines from the point D to the circumference, one upon each side of the least: At the point M, in the straight line MD, make the angle DMB equal to the angle DMK, and join DB: And because MK is equal to MB, and MD common to the triangles KMD, BMD, the two sides KM, MD are equal to the two BM, MD; and the angle KMD is equal to the angle BMD; therefore the base DK is equal to the base DB: But besides DB, no straight line can f 4. 1. be 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 also to DB; therefore DB is equal to DN, that is, the line nearer to DG the least, is equal to the more remote, which is impossible. « If, therefore, any point,” &c. Q. E. D. straight and join Diberiangles KM, MD; one base DK in T! FLEELSETET SEKE are equal to triangles KMD, MK is equale Book 111. PROP. IX. THEOR. If a point be taken within a circle, from which there fall more than two equal straight lines upon the cir. cumference, that point is the centre of the circle. Let the point D be taken within the circle ABC, from which there fall on the circumference more than two equal straight 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 F point D, which is not the centre, DG shall be the greatest line from it to the circumference, and DC greater a than DB, and DB than DA: But they are likewise equal, which is impossible : Therefore E is not the centre of the circle ABC: in like manner, it may be demonstrated, that no other point than D is the centre; D therefore is the centre. Wherefore, " if a point be taken,” &c. Q. E. D. a 7. 3. PROP. X. THEOR. One circumference of a circle cannot cut another in more than two points. If it be possible, let the circum- A moints, viz. inhe circle ABC.cause Ell join KB, KG, KF: And because Ell is a the centre of the circle DEF: But K is also the centre Book III. of the circle ABC; therefore the same point is the centre of two circles that cut one another, which is impossible. There. a 9. 3. fore, “one circumference of a circle cannot cut another in more " " than two points." Q. E. D. 3 PROP. XI. THEOR. Feible, as because, that it, both If two circles touch each other internally, the straight line which joins their centres being produced must pass through the point of contact. Let the two circles ABC, ADE 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 straight line which joins the centres F, G, being produced, passes through the point A. For, if not, let it fall otherwise, if I possible, as FGDH, and join A F, AG: And because AG, GF are greater a than FA, that is, than a 20 1. FH, for FA is equal to FH, both being from the same centre; take away the common part FG; there B fore the remainder AG is greater than the remainder GH: But AG is equal to GD; therefore GD is greater than GH, the less than the greater, which is impossible. Therefore the straight line which joins the points F, G cannot fall otherwise than upon the point A, that is, it must pass through A. Therefore « if two circles,” &c. Q. E. D. PROP. XII. THEOR. If two circles touch each other externally, the straight line which joins their centres must 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 straight line which joins the points F, G must pass through the point of contact A. For, if not, let it pass otherwise, if possible, as FCDG, and |