11 Th. If two circles (ABC, ADE), touch one another internally (in A), the straight line which joins their centres, being produced, shall pass through their point of con tact. Argument. Let F, G, be their central points: then, if GF which joins them do not pass through A, it must pass otherwise, as through HDC: join AF, AG. Then FA, FH are equal radii; as also GA, GH (a): but FA is less than FG and GA (b), or FG and GH, that is, than FH. To be equal and less cannot be admitted: therefore F and G cannot be the true centres, which must be in the same straight line with A, the point of contact. Wherefore, if two circles, &c. E Q. E. D. 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 central points: then, if GF which joins them, do not pass through A, it must pass otherwise, as through FCDG: join AF, AG. Then FA, FC are equal radii; as also GA, GD (a): therefore the two FA, GA are equal to the two FC, GD. B 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 (b), 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; 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 (b), and their centres are in the diameter GH (c); therefore GH passes through their point of contact (d), which is neither B, nor D: hence the circles which affect to touch each E 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 K 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; (c) cor. p. 1, 3; (b) p. 2, 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 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 E 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 (ƒ). 2. And if the central distances EF, EG are equal, the chords AB, CD are also equal: for the squares of the equal radii EA, EC are equal; from which taking the equal squares of EF, EG, the remainders will be equal (g), 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. Recite (a) p. 1, 3; (d) ax. 7; (g) ax. 3; (h) ax. 6. Q. E. D. (c) cor. p. 1, 3; (ƒ) def. 4, 3; 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, (b). Argument 1. The one side BC is less than the 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, FĞ, which are bisected in H, K (e); therefore BC is greater than FG. For 3. And, if BC be greater than FG, it is nearer the centre. 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 BC is nearer the centre than FG (ƒ). Wherefore, the diameter is the greatest chord, &c. Q. E. D. Recite (a) p. 12, 1; (d) p. 47,1; (b) def. 15, 1; (c) p. 20, 1; 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, as AC, can be drawn from the point A, at right angles to AB, so as to fall within the circle, draw DC: B Then, because of the equal radii, DAC is an isosceles triangle (a); and the angles DAC, DCA are equal (b); 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 A 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 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 (ƒ). Therefore AF cuts the circle. Wherefore, the straight line drawn, &c. Q. E. D. (c) p. 32, 1; (f) ax. 9. E Cor. The straight line drawn at right angles to the diameter or radius, from its extremity, touches the circle, and only in one point; for it would cut the circle if it should meet it in two, by p. 2, 3: and in the same point only one line can touch the circle. 17 P. To draw a straight line from a given point (A), without a circle (BCD), or in the circumference (as at D), which shall touch the circle. Constr. Find the centre E (a); join EA; and upon E, with the radius EA, describe the circle AFG; from the point D draw DF at right angles to EA (b); join EBF and AB. Argument 1. Because of the equal radii (c) the triangles AEB, FED have two sides AE, EB in the one, equal to two sides FE, ED in the other, and the angle at E common; therefore (d) the remaining sides AB, FD are equal; as also the angles opposite to the equal sides; namely, ABE to FDE: but FDE is a right angle; and so, ABE is the same; and the radius EB meets AB at right angles in the circumference: therefore AB touches the circle (e), and it is drawn from the given point A. 2. If the given point be in the circumference, as at D, draw DE to the centre, and DF at right angles to DE: DF is drawn to touch the circle (e); which was to be done. Recite (a) p. 1, 3; (d) p. 4, 1; (b) p. 11, 1: (c) def. 15, 1; (e) cor. 16, 3. 18 Th. If a straight line (DE), touch a circle (ABC), the radius (FC) drawn to the point of contact (C) shall be perpendicular to the tangent, or touching line. Constr. Find the centre F (a); and if FC be not perpendicular to DE, draw FBG perpendicular to it (b). Argument. Because FGC affects to be a right angle, FCG must be less than a right angle (c); and FG is therefore less than FC (d), or its equal FB (e); but the whole is greater than its part (f): wherefore FG is not less than FB, nor is it perpendicular to DE; neither is any other line drawn from the centre which does not meet DE in the point of contact. FC is therefore perpendicular to DE. B C CE Q. E. D. Recite (a) p. 1,3; (b) p. 12, 1; (e) def. 15, 1; (c) p. 32, 1; 19 Th. If a straight line (DE) touch a circle (ABC), and from the point of contact (C) a straight line (AC) be drawn in the circle, at right angles to that tangent, the centre of the circle shall be in that straight line. Argument. If the centre of the circle be not in AC, it must be out of it, as in F; join CF. Then because DE touches the circle, and FC affects to be drawn from the centre to the point of contact, FC is perpendicular to DE (a), and the angle FCE is a right angle (b): but ACE is a right angle; and all right angles are equal to one another (c); therefore FCE is equal to ACE: but the whole is greater than its part (d); and two magnitudes cannot be at once equal and unequal: therefore F is not the centre; neither is any point on the right or left of AC. Therefore, if a straight line, &c. D Q. E. D. E Recite (a) p. 18, 3; (c) ax. 10; (b) def. 10, 1; (d) ax. 9. 20 Th. The angle at the centre of a circle is double the angle at the circumference upon the same arc. The angles AEB, BEC, AEC are central: The angles ADB, BDC, ADC are at the circumfer ence, and upon the same arcs as the former. the diameter DEG. Draw Because of the equal radii, the triangles EAD, EBD, ECD, are isosceles (a), and have equal angles opposite to the equal sides (b). And, because the side DE is produced, the exte B rior angle AEG is equal to the two interior opposite angles EAD, EDA (c), or double EDA; the exterior BEG to the two interior EBD, EDB, or double EDB; and the exterior CEG to the two interior ECD, EDC, or double EDC. Therefore, the sum of AEG and BEG is double the sum of ADG and BDG; that is, AEB is double ADB. And the difference of AEG and CEG is double the difference of ADG and CDG; that is, AEC is double ADC. Wherefore, the angle at the centre, &c. Recite (a) def. 24, 1; (b) p. 5, 1; (c) p. 32, 1. 21 Th. Q. E. D. The angles in the same segment of a circle are equal to one another. Let BAED be a segment: the angles BAD, BED are equal to one another. 1. If the centre F be in the segment, join FB, FD: then because the angle BFD is at the centre and the angles BAD, BED at the circumference. on the same arc, each of the latter is half the former (a); and halves of the same are equals (b): there. fore BAD is equal to BED. |