FEB meeting them in E and B; and because F is the centre of the circle ABC, CF is equal to FB: Also, because F is the centre of the circle CDE, CF is equal to FE: And CF was shewn to be equal to FB; therefore FE is equal to FB, the less to the greater, which is impossible: Wherefore F is not the centre of the circles ABC,CDE. Therefore if two circles, &c. Q. E. D. Proposition VII. Theorem. A If any point be taken in the diameter of a circie 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 which passes 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. B Join BE, CE, GE; and because two sides of a triangle are greater (20. 1.) than the third, BE, EF are greater than BF; but AE is 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 (24. 1.) than the base FC: For the same reason, CF is greater than GF; Again, because GF, FE are greater (20. 1.) than EG, and EG is equal to ED; GF, FE are greater than ED : C 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 shortest line FD: At the point E in the straight line EF, make (23. 1.) the angle FEH equal to the angle GEF and join 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 angle HEF; therefore the base FG is equal (4. 1.) to the base FH But besides FH, no other straight line can de arawn 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; tha nearer to that which passes through the centre equal to one is more remote; which is impossible. Therefore, if any be taken, &c. Q. E. D. 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 s 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 nearer to it 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 (1. 3.) M the centre of the circle ABC, and join ME, MF, MC, MK, ML, MH: And because AM is equal to ME, add MD to each, therefore AD is equal to EM, MD; but EM, MD are greater (20. 1.) than ED; therefore also AD is greater than ED: Again, because ME is equal to MF, and MD common to the triangles EMD, FMD; EM, MD are equal to FM, MD; but M the angle EMD is greater than the angle FMD; therefore the base ED is greater (24. 1.) than the base FD In like manner it may be shewn 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 (20. 1.) than MD, and MK is equal to MG, the remainder KD is greater (4 Ar.) than the remainder GD, that is, GD is less than KD: And because MK, DK are drawn to the point K within the triangle MLD from M, D, the extremities of its side MD, MK, KD are less (21. 1.) than ML, LD, whereof MK is equal to ML; therefore the remainder DK is less than the remainder DL: In like inanner it may be shewn, 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 (4. 1.) to the base DB: But, besides DB, there can be no straight 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 also to DB; therefore DB is equal to DN, that is the nearer to the least equal to the more remote, which is impossible. If, therefore, any point, &c. Q. E. D. If a point be taken within a circle, from which there fall more than two equal straight 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 straight lines, viz. DA, DB, DC, the point D is the centre of the circle. DE F G 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, DG shall be the greatest line from it to the circumference, and DC greater (7. 3.) 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 but D is the centre; D therefore is the centre. Wherefore, if a point be taken, &c. Q. E. D. Proposition X. Theorem. B One circumference of a circle cannot cut another in more than two points. E B H D K F C G If it be possible, let the circumference FAB 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 straight lines KB, KG, KF, the point K is (9. 3.) the centre of the circle DEF: But K is also the centre of the circle ABC; therefore the same point is the centre of two circles that cut one another, which is impossible. (5. 3.) Therefore one circumference of a circle cannot cut another in more thar. two points. Q. E. D. If two circles touch each other internally, the straight line which joins their centres being produced, shall 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. H GE E For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG: And because AG,GF are greater (20. 1.) than FA, that is than FH, for FA is equal to FH, both being from the same B centre; take away the common part FG; therefore 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 it. Therefore, if two circles, &c. Q. E. D. Proposition XII. Theorem. If two circles touch each other externally, the straight line which joins their centres, shall 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 shall pass through the point of contact A. E For, if not, let it pass otherwise, if possible, as FCDG, and join FA, AG; and because F is the centre of the circle ABC, AF is equal to FC: Also because G is the centre of the circle ADE, AG is equal to GD: Therefore FA, AG are equal to FC, DG: wherefore the whole FG is greater than FA, AG: But it is also less; (20.1.) which is impossible: Therefore the straight line which joins the points F, G shall not pass otherwise than through the point of contact A, that is, it must pass through it. Therefore, if two circles, &c. Q. E. D. Proposition XIII. Theorem. F G One circle cannot touch another in more points than one, whether it touches it on the inside or outside. For, if it be possible, let the circle EBF touch the circle ABC in more points than one, and first centres are (Cor. 1. 3.) in the straight line GH which bisects BD at right angles: Therefore GH passes through the point of contact; (11.3.) but it does not pass through it, because the points B, D are without the straight line GH, which is absurd: Therefore one circle cannot touch another on the inside in more points than one. Nor can two circles touch one another on the outside in more than one point: For, if it be possible, let the circle ACK touch the circle ABC in the points A, C, and join AC: Therefore, cause the two points A, C, are in the circumference of the circle ACK, the straight line AC which joins them shall fall within (2. 3.) the circle ACK: and the circle ACK is without the circle ABC; and therefore the straight line AC is without this last circle; but because the points A, C are in the circumference of the circle ABC, the straight line AC must be within (2. 3.) the same circle, which is absurd: therefore one circle cannot touch another on the outside in more than one point: And it has been shewn, that they cannot touch on the inside in more points than one. Therefore, one circle, &c. Q. E. D. Proposition XIV. Theorem. B Equal straight lines in a circle are equally distant from the centre; and those which are equally distant from the centre, are equal to one another. Let the straight lines AB, CD, in the circle ABDC, be equal to one another: they are equally distant from the centre. Take E the centre of the circle ABDC, and from it draw EF, EG perpendiculars to AB, CD: Then because the straight line EF, passing through the centre, cuts the straight line AB, which does not pass through the centre at right angles, it also bisects (3. 3.) it: Wherefore AF is equal to FB, and AB double of AF. For the same 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 square of AE is equal to the square of EC: But the squares of AF, FE are equal (47. 1.) to the square of AE, because the angle AFE is a right angle; and for the like reason, the squares of EG, GC, are equal to the square of EC: B Therefore the squares, of AF, FE are equal to the squares of CG, GE, of which the square of AF is equal to the square of CG, because AF is equal to CG; therefore the remaining square of FE is equal to the remaining square of EG, and the straight line FE is therefore equal to EG: But straight lines in a circle are said to be equally distant from the centre, when the perpendiculars drawn to them from the centre are equal: (4. Def. 3.) Therefore AB, CD are equally distant from the centre. E Next, if the straight lines AB, CD be equally distant from the centre, that is, if FE be equal to EG, AB is equal to CD: For the same construction being made, it may, as before, be demonstrated, |
