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Argument 1. If straight lines be drawn from
BA E and F in the diameter, to the points B, C, G in the circumference; there will be in each of the triangles BEF, CEF, GEF, a radius, and the side EF common, which are equal to AF: but one side of a triangle is less than the sum of the other two (a); therefore FG, FC, or FB is less than FA, which passes through the centre.
2. And, because EG, or its equal ED (6), that is EF, FD, is less than EF, FG (a), taking away
Ο Η. the common part EF, the remainder FD is less than the remainder FG; but FD is the other part of the diameter.
3. Also, since the sides CE, EF, though equal to the sides BE, EF, contain a less angle; therefore (c) the base CF is less than the base BF. For like reason the base GF is less than the base CF (d).
4. Make now the angle FEH equal to the angle FEG; join FH: then because EG, EH are equal radii
, and EF is common; the bases FG, FH are equal (e). But, besides FH, no straight line can be drawn from the point F to the circumference equal to FG; for such line, as FK, must incline towards the greatest or least, and be greater or less than FH or FG.
Therefore from any point, not the centre, &c. Q. E. D.
(d) p. 23, 1; (e) p. 4, 1.
8 Th. From any point (A) without a circle (BCD), the greatest straight line (AB) that can be drawn to the con. cave arc passes through the centre (M); and other lines, drawn from the same point, diminish as they decline from the greatest. But of those which meet the convex arc, the least (AE) is the exterior part of the greatest; and the rest increase as they decline from the least: and only two of them can be equal, one on each side of the least.
Argument i. The distance from A through M to B, C, or D is equal; but the straight line AC is less than the distance from A to C through M (a); therefore ABC is greater than AČ, and it passes through the centre M.
2. Because the angle AMD is less than the angle AMC, the base AD is less than the base AC (6), and it is more remote from AB.
3. Again, because AE and EM are less than AF and FM (a), and that the radii EM and FM are equal, AE is less than AF, and it is the exterior part of AB.
4. Also, since AF, FM are less than AG, GM (c); take away the equal radii FM, GM, the remainder AG is greater than AF, and it is more remote from AE.
5. Make now the angle AMH equal to the angle AMF (d), and join AH: then in the triangles AFM, AHM, the side AM is common; HM and FM are equal radii
, and the angles AMH, AMF are equal; therefore these two triangles are equal, and have the side AH equal to AF (e). Moreover, no line but AH can be drawn to the circumference, equal to AF: for such line, as Al, must decline more or less from AE, and be greater or less than AF.
Wherefore, from any point without a circle, &c. Q. E. D,
(d), p. 23, 1; (e) p. 4, 1.
9 Th. If from a point (D), within a circle (ABC), more than two equal straight lines can be drawn, that point is the centre of the circle.
Constr. From the point D draw three equal straight lines DA, DB, DC: then, if d be not the centre, let it be E; and, through E and D, draw FG; which is therefore a diameter of the circle (a).
Argument. Because D is a point in the diameter of a circle, not the centre, DG, which passes through the centre, is the greatest straight line that can be drawn from it to the circumference (b); and of the rest, DC is greater than DB, and DB than DA: but these three were assumed to be equal; therefore the conditions are inconsistent; and must be, while È, or any point except D, is taken as the Wherefore, if from a point, &c.
Q. E. D. Recite (a) def. 16, 1; (b) p. 7, 3.
10 Th. One circumference of a circle cannot cut another in more than two points.
If it be possible, let the circumferences ABC,
A there are drawn to the circumference more than two equal straight lines: the point K is therefore the centre of the circle DEF (a), as well as that of ABC: and so, two circles which cut one another have the same centre (b); which cannot be said of true circles..
Wherefore, one circumference of a circle, &c. Q. E. D.
11 Th. If two circles (ABC, ADE), touch one another internally in A), the straight line which joins their cen. tres, being produced, shall pass through their point of contact.
Argument. Let F, G, be their central points :
Then FA, FH are equal radii; as also GA, GH
Q. E. D. Recite (a) def. 15, 1; (6) p. 20, 1.
12 Th. If two circles (ABC, ADE), touch each other externally (in A), the straight line which joins their centres shall pass through the point of contact. Argument. Let F, G, be their cen
B tral points: then, if GF which joins them, do not pass through A, it must pass otherwise, as through FCDG: join AF, ÁG.
Then FA, FC are equal radii; as also GA, GD (a): therefore the two FA, GA are equal to the two FC, GD. But FG is greater than FC, GD, and therefore greater than FA, GA; and so, one side of a triangle is greater than the other two; but it is also less (6), which is impossible. Therefore F and G are not the true centres, which must be in the same straight line with A, the point of contact. Wherefore, if two circles, &c.
Q. E. D. Recite (a) def. 15, 1; (b) p. 20, 1.
13 Th. One circle cannot touch another in more than one point, whether they touch inside or outside.
Argument 1. - If the circles ABC, DEF are said to touch each other internally in two points B, D;
B join BD; and draw GH to bisect BD at right angles (a.) Then, as the points B, D, are in the circumference of each of the circles, the chord BD falls within both (6), and their centres are in the.
other in B, D, are not true; and so, one circle cannot touch another on the inside in more than one point.
2. Again, let the circles ABC, ACK touch each other externally, as in A, C; join AC. Then because the points A, C, are in the circumference of each of the circles, the chord AC falls within both (6); and so, while affecting only to touch, the circles intersect each other: therefore one circle cannot touch another on the outside in more than one point. Wherefore, one circle cannot touch, &c.
Q. E. D. Recite (a) p. 10, 11, 1; (6) p. 2, 3;
(c) cor. p. 1, 3; (d) p. 11, 3.
14 Th. Equal chords (AB, CD), in a circle (ABCD), are equidistant from the centre; and chords equidistant from the centre are equal to one another.
Constr. Find the centre E (a), from which draw perpendiculars EF, EG, upon AB, CD (6): join EA, EC.
Argument 1. The equal chords AB, CD are bisected by the perpendiculars EF, EG, drawn from E the centre (d). Wherefore, their halves AF, CG are equal (d); and the squares of the halves are equal. But the squares of AF, FE, and of CG, GE, are equal to the squares of the equal radii EA, EC, each to each: therefore the squares of AF, FE are equal to the squares of CG, GE(e): from these take the equal squares of AF, CG; the remaining squares FE, GE are equal: and so, EF is equal to EG: therefore the chords AB, CD are equidistant from the centre (J).
2. And if the central distances EF, EG are equal, the chords AB, CD are also equal: for the squares of the equal radü EA, EC are equal; from which taking the equal squares of EF, EG, the remainders will be equal (8), which are the squares of AF, CG: therefore AF is equal to CG; and they are halves of AB, CD, which are therefore equal (h). Wherefore, equal chords in a circle, &c.
Q. E. D. Recite (a) p. 1, 3;
(b) p. 12, 1, and 3, 3; (c) cor. p. 1, 3; (d) ax. 7; (e) p. 47, 1;
(f ) def. 4, 3; (g)ax. 3 ; (h) ax. 6.
15 Th. The diameter (AD) is the greatest chord in a circle (ABCD); and of all others, the chord (BC) nearer to the centre (E) is greater than one more remote (FG); and the greater is nearer to the centre than the less.
Constr. Draw EH, EK perpendicular on BC, FG (a); join EB, EC, EF,(6).
Argument 1. The one side BC is less than the
BA two EB, EC (c), or their equals EA, ED (6), that is, AD, which is therefore greater than BC.
2. The equal radii EB, EF have equal squares, and they subtend right angles at H, K (d); therefore the two squares on EH, HB and the two on EK, KF are equal. But EH is less than EK, and its square is the less; therefore the square on HB exceeds that on KF; and so HB is greater than KF. But these are the halves of BC, FG, which are bisected in H, K (e); therefore BC is greater than FG.
3. And, if BC be greater than FG, it is nearer the centre. For the two squares on EH, HB prove equal to the two on EK, KF, and that on HB exceeds that on KF; therefore, of the two squares which remain, that on EH is less than that on EK; and so, EH is less than EK: therefore BĆ is nearer the centre than FG (). Wherefore, the diameter is the greatest chord, &c.
Q. E. D. Recite (a) p. 12, 1; (6) def. 15, 1; (c) p. 20, 1;
(d) p. 47,1; (e) p. 3, 3; (f) def. 4, 3.
16 Th. The straight line (AE) drawn at right angles to the diameter (AB) of a circle, from its extremity, falls without the circle; and no straight line can make an acute angle with that diameter so great, or with the line which falls without the circle so small, as not to cut the circle.
1. For if a straight line, às AC, can be drawn from the point A, at right angles to AB, so as to fall within the circle, draw DC:
Then, because of the equal radii, DAC is an isosceles triangle (a); and the angles DAC, DCA are equal (6); and both right angles: for DAC is a right angle: add the angle ADC; therefore the three angles of a triangle are greater than two right angles (c), which is impossible. Therefore a straight line cannot be drawn from the point À to fall within the circle, and be at right angles with the diameter. Neither can it fall upon the circumference and be a straight line (d). It falls therefore without the circle. 2. Again, if a straight line, as AF, can be drawn
F. E between AE and the circumference, make DG perpendicular to AF, cutting the circumference in H: then, because DGA affects to be a right angle, and DAG is acute, the side DA (e), or its equal DH,
B must be greater than DG: but a part is not greater than the whole (f). Therefore AF cuts the circle. Wherefore, the straight line drawn, &c.
Q. E. D. - Recite (a) def, 24, 1; (6) p. 5, 1; (c) p. 32, 1;
(d) def. 4,5, 1; (e) p. 18, 1; (f) ax. 9,