Proposition 5.-Theorem. If two circles cut one another, they shall not have the same centre. Let the two circles ABC, CDG cut one another in the points B, C. They shall not have the same centre. For, if it be possible, let E be their centre; join EC, and draw any straight line EFG, meeting the circles in F and G. PROOF.-Because E is the centre of the A circle ABC, EC is equal to EF (I. Def. 15). And because E is the centre of the circle CDG, EC is equal to EG. But EC was shown to be equal to EF; Therefore EF is equal to EG (Ax. 1), the less to the greater, ..EF=EC, which is impossible; Therefore E is not the centre of the circles ABC, CDG. Proposition 6.-Theorem. If one circle touch another internally, they shall not have the same centre. Let the circle CDE touch the circle ABC internally in the point C. They shall not have the same centre. CONSTRUCTION,--For, if it be possible, let F be their centre; If they have join FC, and draw any straight line FEB, meeting the circles in E and B. PROOF.-Because F is the centre of the circle ABC, FC is equal to FB (I Def. 15). And because F is the centre of the circle CDE, FC is equal to FE. But FC was shown to be equal to FB; the same centre F, B F E FC-FE and FB. Therefore FE is equal to FB (Ax. 1), the less to the greater, ...FE=FD which is impossible; Therefore F is not the centre of the circles ABC, CDE, FA> FB Proposition 7.-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 this point 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 straight line which passes through the centre is always greater than one more remote; and from the same point there can be drawn to the circumference two straight lines, and only two, which are equal to one another, one on each side of the shortest line. Let ABCD be a circle, AD its diameter, and E its centre, in which let any point F be taken which is not the centre. Of all the straight lines FB, FC, FG, &c., that can be >FCFG drawn from F to the circumference, FA, which passes &c., >FD the least. through the centre, shall be the greatest; For or AF BF And base BF FD, the other part of the diameter AD, shall be the least; And of the others, FB, the nearer to FA, shall be greater than FC, the more remote; and FC greater than FG. CONSTRUCTION.—Join BE, CE, GE. PROOF.-Because any two sides of a triangle are greater than the third side, BE, EF are greater than BF (I. 20). A B D H But AE is equal to BE; therefore AE, EF, that is, AF is greater than BF. Again, because BE is equal to CE, and EF common to the two triangles BEF, K CEF, the two sides BE, EF are equal to the two CE, EF, each to each. But the angle BEF is greater than the angle CEF; > base FC (I. 24). Again ..GF>FD In the same manner it may be shown that FC is greater than FG. Again, because GF, FE are greater than EG (I. 20), and that EG is equal to ED; Therefore GF, FE are greater than ED. Take away the common part FE, and the remainder GF is greater than the remainder FD (Ax. 5). Therefore FA is the greatest, and FD the least, of all the straight lines from F to the circumference; and FB is greater than FC, and FC than FG. Also, there cannot be drawn more than two equal straight lines from the point F to the circumference, one on each side of the shortest line. CONSTRUCTION.-At the point E, in the straight line EF, make the angle FEH equal to the angle FEG (I. 23), and join FH. GEF PROOF.-Because EG is equal to EH, and EF common to Triangles the two triangles GEF, HEF, the two sides EG, EF are equal to the two sides EH, EF, each to each; And the angle GEF is equal to the angle HEF (Const.); For, if it be possible, let FK be equal to FG; Proposition 8.--Theorem. im If any point be taken without a circle, and straight lines be drawn from it to the circumference, one of which passes through the centre; of those which fall on the concave circumference, the greatest is that which passes through the centre, and of the rest, that which is nearer to the one passing through the centre is always greater than one more remote; but of those which fall on 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 one more remote; and from the same point there can be drawn to the circumference two straight lines, and only two, which are equal to one another, one on each side of the least line. Let ABC be a circle, and D any point without it, and from D let the straight lines DA, DE, DF, DC be drawn and HEF are equal in every respect. And if FK =FG, it also = FH, which is impossible. DA DE and DG DK <DL DH to the circumference, of which DA passes through the centre. Of those which fall on the concave part of the circumference AEFC, the greatest shall be DA, which passes through the centre, and the nearer to it shall be greater than the more remote, viz., DE greater than DF, and DF greater than DC. But of those which fall on the convex circumference GKLH, the least shall be DG between the point D and the diameter AG, and the nearer to it shall be less than the more remote, viz., DK less than DL, and DL less than DH. CONSTRUCTION.-Take M, the centre of the circle ABC (III. 1), and join ME, MF, MC, MH, ML, MK. For K GB Also base ED &c. Again M N PROOF.-Because any two sides of a triangle are greater than the third side, EM, MD are greater than ED (I. 20). But EM is equal to AM; therefore AM, MD are greater than ED—that is, AD is greater than ED. Again, because EM is equal to FM, and MD common to the two triangles EMD, FMD; the two sides EM, MD are equal to the two sides FM, MD, each to each; But the angle EMD is greater than the angle FMD; Therefore the base ED is greater than the base FD (I. 24). 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 greater than DC. Again, because MK, KD are greater than MD (L. 20), >MD, or and MK is equal to MG; >MK+ The remainder KD is greater 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 and D, the extremities of its side MD; Therefore MK,DK are less than ML, LD (I. 21). But MK is equal to ML; therefore the remainder KD is less than the remainder LD. In like manner it may be shown that LD is less than HD. Therefore DG is the least, and KD less than DL, and DL less than DH. Also, there can be drawn only two equal straight lines from the point D to the circumference, one on each side of the least line. CONSTRUCTION.-At the point M, in the straight line MD, make the angle DMB equal to the angle DMK (I. 23), and join DB. KMD are equal in every PROOF.-Because MK is equal to MB, and MD common Triangles to the two triangles KMD, BMD; the two sides KM, MD and BMD are equal to the two sides BM, MD, each to each; And the angle DMK is equal to the angle DMB (Const); respect. Therefore the base DK is equal to the base DB (I. 4). But, besides DB, no other straight line can be drawn from D to the circumference equal to DK. For, if it be possible, let DN be equal to DK. Then, because DN is equal to DK, and DB is also equal DN=DK to DK, therefore DB is equal to DN (Ax. 1); and.. =DB. impossible. That is, a line nearer to the least is equal to one which is which is more remote; which is impossible by what has been already shown. Therefore, if any point, &c. Q.E.D. Proposition 9.-Theorem. If a point be taken within a circle, from which there can be drawn 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 can be drawn more than two equal straight lines, viz., DA, DB, DC. The point D shall be the centre of the F circle. CONSTRUCTION.-For if not, let E be the centre; join DE, and produce it to the circumference in F and G. A DE G |