DA DE and DG DK <DL DH For Also Again MK+KD >MD, or >MK+ DG. ...KD> DG, and..DG <KD, &c. 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. GB M 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 (I. 20), and MK is equal to MG; The remainder KD is greater than the remainder GDthat 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 circle. CONSTRUCTION. For if not, let E be the centre; join DE, and produce it to the circumference in F and G. A DE If D be not PROOF. Then FG is a diameter of the circle ABC (I.Def. 17). the centre. is taken the point D, not the centre; DGDC But they are also equal. Therefore DG is the greatest straight line from D to the circumference, and DC is greater than DB, and DB greater than DA (III. 7); But these lines are likewise equal, by hypothesis; which is impossible. Therefore E is not the centre of the circle ABC. In like manner it may be demonstrated that any other point than D is not the centre; Therefore D is the centre of the circle ABC. Therefore, if a point, &c. Q.E.D. Proposition 10.-Theorem. One circumference of a circle cannot cut another in more than two points. CONSTRUCTION. -If it be possible, let the circumference If possible, B E the two circles have the same centre. ABC cut the circumference DEF in more than two points, viz., in the points B, G, F. Take the centre K of the circle ABC (III. 1), and join KB, KG, KF. PROOF.-Because K is the centre of the circle ABC, the radii KB, KG, KF are all equal. 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, therefore K is the centre of the circle DEF (III. 9). But K is also the centre of the circle ABC (Const.); Therefore the same point is the centre of two circles which cut one another; which is impossible (III. 5). Therefore, one circumference, &c. Q.E.D. Proposition 11.-Theorem. If one circle touch another internally in any point, the straight line which joins their centres, being produced, shall pass through that point. Let the circle ADE touch the circle ABC 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 their centres, being produced, shall pass through the point of contact A. CONSTRUCTION.-For, if not, let it pass otherwise, if If not, possible, as FGDH. Join AF and AG. PROOF.-Because AG, GF are greater than AF (I. 20), and AF is equal to HF (I. def. 15); Therefore AG, GF are greater than HF. Take away the common part GF, and the remainder AG is greater than the remainder HG. H G AG-IIG. E B But AG But AG is equal to DG (I. Def. 15); Therefore DG is greater than HG, the less than the DG. greater; which is impossible. Therefore the straight line which joins the centres, being produced, cannot fall otherwise than upon the point A, that is, it must pass through it. Therefore, if one circle, &c. Q.E.D. Proposition 12.-Theorem. If two circles touch each other externally in any point, the straight line which joins their centres shall pass through that point. 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 the circle ADE; The straight line which joins their centres shall pass through the point of contact A. CONSTRUCTION.-For, if not, let it pass otherwise, if possible, as FCDG. Join FA and AG. PROOF.-Because F is the centre of the circle ABC, EA is equal to FC (I. Def. 15). And because G is the centre of the circle ADE, GA is equal to GD; ..DG> HG. If not, FG is> FA+AG, but it is Therefore FA, AG are equal to FC, DG (Ax. 2). But FG is also less than FA, AG (I. 20), which is imalso less. possible. If possible, Therefore the straight line which joins the centres of the circles shall not pass otherwise than through the point A, that is, it must pass through it. Therefore, if two circles, &c. Q.E.D. Proposition 13.—Theorem. One circle cannot touch another in more points than one, whether it touch it internally or externally. I. First, let the circle EBF touch the circle ABC internally in the point B. Then EBF cannot touch ABC in any other point. CONSTRUCTION.-If it be possible, let EBF touch ABC in let it touch another point D; join BD, and draw GH bisecting BD at right angles (I. 10, 11).. in D also; then PROOF.-Because the two points B, D are in the circumference of each of the circles, the straight line BD falls within each of them (III. 2). Therefore the centre of each circle is in the straight line GH, which bisects BD at right angles (III. 1 cor.) Therefore GH passes through the point of contact (III. 11). But GH does not pass through the point of contact, the point because the points B, D are out of the line of GH; which GH passes through ofcontact, is absurd. which it does not. Therefore one circle cannot touch another internally in more points than one. |