another AC which does not pass through the centre, cut it at right angles (III. 3); FEA is a right angle. wherefore Again, because the straight line FE bisects the straight line BD, which does not pass through the centre, and FEA was it shall cut it at right angles; it shall wherefore therefore therefore AC and BD do If two circles cut one another, they shall not have the same centre. Given the two circles ABC and CDG, which cut one another in the points B, C; to prove that they have not the same centre. (Const.) For, if it be possible, let E be their centre; join EC, and draw any straight line EFG meeting them in F and G; (Dem.) and because E is the centre of the circle ABC, CE is Lequal to EF. Again, because E is the centre of the circle CDG, CE is equal to EG; but CE was shewn to be equal therefore EF is equal to EG, to EF; A the less to the greater, which is impossible; is not the centre of the circles ABC and CDG, may be proved of any other point, same centre. hence they have not the PROPOSITION VI. THEOREM. If one circle touch another internally, the circles shall not have (Const.) For, if they can, let it be F; join FC, and draw any straight line FEB meeting them in E and B; (Dem.) and because F is the centre of the circle ABC, also, because F CF is equal to FB; is the centre of the circle CDE, CF is equal to FE; and CF was shewn equal to FB; therefore FE is equal to FB, greater, which is impossible; the circles ABC and CDE, and the same may be proved of any other point, hence they have not the same centre. PROPOSITION VII. THEOREM. 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 the 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 cannot be drawn more than two straight lines that are equal to one another, one upon each side of the diameter. Given a circle ABCD, and AD its diameter, and E the centre, in which any point F is taken which is not the centre. To prove that 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. (Dem.) and (I. 20) BE, EF are therefore AE, (Const.) Join BE, CE, GE; greater than BF; but AE is equal to EB; EF, that is, AF is greater than BF. Again, because BE is equal to CE, and FE common to the triangles BEF and CEF, the two sides BE and EF are equal to the two CE and EF; but the angle BEF is greater than the angle CEF; therefore the base BF is greater than the base FC (I. 24); like reason, CF is greater than GF. for a Again, because GF and FE are greater than EG, and EG is equal to ED; GF and FE are and there greater than ED; take away the common part FE, the remainder GF is greater than the remainder FD; fore 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 cannot be drawn more than two equal straight lines from the point F to the circumference, one upon each side of make the diameter; at the point E in the straight line EF, the angle FEH equal to the angle GEF (I. 23), and join FH. Then, because GE is equal to EH, and EF common to the two triangles GEF and HEF; are equal to the two HE and EF; the two sides GE and EF and the angle GEF is equal to the angle HEF; therefore the base FG is equal to the base FH (I. 4). 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; FK is equal to FH; and because FK is equal to FG, and FG to FH, that is, a line nearer to that which passes through the centre, is equal to one which is more remote, which is impossible. 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 to the circumference, one upon each side of the least. Given a circle ABC, and D any point without it, from which are drawn the straight lines DA, DE, DF, and DC to the circumference, whereof DA passes through the centre. To prove that 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 ; namely, DE greater than DF, and DF greater 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; namely, DK less than DL, and DL less than DH. (Const.) Take M the centre of the circle ABC (III. 1), and join ME, MF, MC, MK, ML, and MH. (Dem.) And because AM is equal to ME, add MD to each; therefore AD is equal to EM and but EM and MD are greater than MD; ED (I. 20); therefore, also, AD is greater than ED. to FM and MD; D M C F E A Again, because ME is equal to MF, and MD common to the triangles EMD and FMD; EM and MD are equal but the angle EMD is greater than the therefore the base ED is greater than the base 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 than MD (I. 20), and MK is equal to MG, the remainder KD is greater than the remainder GD that is, GD is less than KD. (Ax. 5); And because MK, DK are drawn to the point K within the triangle MLD from M and D, the extremities of its side MD; MK and KD are less than ML and LD (I. 21), whereof MK is equal to ML; therefore the remainder DK is less than the remainder DL. In like manner 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; for at the point M, in the straight line MD, make the angle DMB equal to the angle DMK (I. 23), and join DB; and, because in the triangles KMD and BMD, side KM is equal to the side BM, and MD common to both, the and also the angle KMD equal to the angle BMD, the base DK is equal to the base DB (I. 4). straight line can be drawn from D to the DK; for, if there can, let it be DN; equal to DK, and DK equal to DB, But, besides DB, no circumference, equal to then, because DN is DB is equal to DN; that is, the line nearer to DG, the least is equal to the more remote, which has been shewn to be impossible. 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. Given the point D within the circle ABC, from which to the point D is the the circumference there fall more than two equal straight linesnamely, DA, DB, and DC; to prove that centre of the circle. (Dem.) For, if not, let E be the centre; join DE, and produce it to the circumference in F and G; diameter of the circle ABC. then FG is a F A 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 than DB (III. 7), 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. Schol.-The proposition is easily proved without a figure, thus: The point from which more than two equal lines can be drawn to the circumference of a circle, cannot be an eccentric point (III. 7); therefore it must be the centre. The circumference of a circle cannot cut that of another in more than two points. (Const.) If it be possible, let the circumference FAB cut the circumference DEF in more than two points; namely, in B, G, and F; centre K of the circle ABC, take the and join and KB, KG, and KF. (Dem.) And because from which to the circumference DEF Ε the fall more than two equal straight lines KB, KG, and KF, point K is the centre of the circle DEF (III. 9); 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 (III. 5). If two circles touch each other internally, the straight line which joins their centres being produced, shall pass through the point of contact. Given the two circles ABC, ADE touching 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; to prove that the straight line which joins the centres F, G, being produced, passes through the point A. (Dem.) For, if not, let it fall otherwise, if possible, as FGDH, and join AF, AG; and because AG, GF are greater than FA (I. 20); that is, than FH, for FA is equal to FH, being each a radius of the same circle; take away the common part FG, and the remainder AG will be greater than the remainder GH. But AG is equal to GD, being each a radius of the circle ADE, therefore GD is greater than GH; is also less, which is impossible; and it therefore HD the straight line which joins the centres cannot fall otherwise than on the point A; that is, it must pass through A. |