BOOK THIRD. Definitions. 1. Equal circles are those of which the diameters or radii are equal. 2. A tangent is a straight line touching a circle, without cutting it, however produced. 3. A circle touches a circle when they meet and do not cut one another. 4. Chords are equidistant from the centre, when the perpendiculars drawn to them from the centre are equal: when the perpendicular is greater the chord is more remote. arc. 5. The angle of a segment is the declination of its chord from the 6. An angle is in a segment when the sides containing it are in the segment: and an angle is said to insist, or stand upon the arc intercepted between the sides containing the angle. 7. Similar segments of a circle are those which contain equal angles. 10802 Propositions. 1 P. To find the centre of a given circle (ABC). Construction. In the circle draw any chord AB, and bisect it in D (a); through D draw CE at right angles to AB (6); bisect CE in F (a); draw FA, FB (c). F D B Argument. In the triangle ADF, BDF, the sides AD, BD are made equal, and DF is common; also the angles at D are equal, being right angles (d): there- A fore the sides FA, FB are equal (e); and, being drawn from the bisectional point of CE, they are radii of the circle (f) wherefore the point F is the centre sought. Recite (a) p. 10, 1; (b) p. 11, 1; (d) ax. 10, 1; (e) p. 4, 1; (c) post. 1; (f) def. 15, 1. E Corollary. The line which bisects another at right angles in a circle, passes through the centre of a circle. 2 Th. If any two points (A, B) be taken in the circumference of a circle (ABC), the straight line (AB) which joins them, shall fall within the circle. Constr. Find the centre D (a), and draw radii from D to A, B, F; also draw DE perpendicular upon AB (6). D F C Argument. In the triangle DAE the exterior angle DEB exceeds the interior DAE (c), or its equal DBE (d); therefore the radius DB, or its equal DF (e), exceeds the side DE (f); and so the part DF exceeds A E B the whole DE, which cannot be admitted (g). Wherefore AB falls not without the circle; and it cannot fall upon the circumference because it is a straight line (h): it therefore falls within the circle. Recite (a) p. 1, 3; Q. E. D. (d) p. 5, 1; (g) ax. 9, 1; (e) def. 15, 1; (f) p. 18, 1; 3 Th. If a diameter (CD) of a circle (ABC) bisect a chord (AB) not passing through the centre, the former will cut the latter at right angles; and the chord so cut shall be bisected (in F). Argument. Find the centre E (a), and join EA, EB. Then, in the triangles EAF, EBF the bases AF, BF are equal, by hyp.; EA, EB are equal radii, and EF is common: therefore the adjacent angles AFE, BFE are equal (6); and so CD bisects AB at right angles in F (c). C E F B D Again, since CD cuts AB at right angles, the angles A at Fare equal (d); and so are the angles at A and B, which are opposite to equal radii (e); therefore the angles AEF, BEF are equal (f), and likewise the sides AF, FB; that is, the chord AB is bisected in F. Wherefore, if a diameter of a circle, &c. Q. E. D. Recite (a) p. 1, 3; (b) p. 8, 1; (c) p. 13, 1; (d) ax. 10, 1; (e) p. 5, 1; (f) p. 26, 32, or 4 b. 1. 4 Th. If in a circle (ABCD) two chords (AC, BD) cut each other (in E), which do not both pass through the centre, they do not bisect each other. Argument 1. Let F be the centre; then, if AC pass through it, BD cannot pass through F, by hyp.; therefore AC is not bisected by BD (a). 2. If neither of the chords pass through F, draw FE from the centre to the sectional point. Now because FE affects to bisect AC, and also BD, the angles FEA, FEB assume to be right angles (b), and equal to each other (c); and so, a part equals the whole, which is is impossible im (d): therefore neither of the chords is bisected ın E. Wherefore, if in a circle, &c. Recite (a) def. 14, 15, 1; (c) ax. 10, 1; Q. E. D. (b) p. 3, 3; (d) ax. 9, 1. 5 Th. If two circles (ABC, ABG` cut each other, they shall not have the same centre. Argument. For, if possible, let E be the common centre; and let C be one of their sectional points; join CE, and draw a straight line EFG, to meet the circles in F and G. Then EC and EF, also EC and EG, are equal radii (a); and so, EF is equal to EG (6)-a part equal to the whole, which cannot be admitted (c). Wherefore E is not the common centre; and E is any point whatever. Hence, if two circles, &c. Recite (a) def. 15, 1; (6) ax. 1; (c) ax. 9. B A E C E. D. 6 Th. If two circles (ABC, CDE) touch each other internally, they shall not have the same centre. Argument. For, if possible, let F be the common centre, and C the point of contact: join FC, and draw a straight line FEB to meet the circles in E and B. Then FC and FE, also FC and FB are equal radii (a); and so, FE and FB are equal (b)-a part equal to the whole, which is impossible (c). Wherefore, F is not the common centre; nor is any other point. Therefore, if two circles, &c. Recite (a) def. 15, 1; (b) ax. 1; (c) ax. 9. D E B F A C Q. E. D. 7 Th. From any point (F) not the centre, in the diameter (AD) of a circle (ABCD), the greatest straight line drawn to the circumference passes through the centre (E) : and FD, the other part of the diameter, is the least: and other lines drawn from that point diminish as they recede from the greatest towards the least: also from the same point, only two equal straight lines can be drawn, one on each side of the diameter. Argument 1. If straight lines be drawn from 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. C BA K 2. And, because EG, or its equal ED (6), that is EF, FD, is less than EF, FG (a), taking away the common part EF, the remainder FD is less than the remainder FG; but FD is the other part of the diameter. CDH 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. (c) p. 24, 1; 8 Th. From any point (A) without a circle (BCD), the greatest straight line (AB) that can be drawn to the concave 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 1. 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 AB C is greater than AC, and it passes through the centre M. B 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. D M A 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 AI, 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, Recite (a) p. 20, 1; (d), p. 23, 1; (6) p. 24, 1; (c) p. 21, 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. C B A ED F 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 diame- C ter 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 E, or any point except D, is taken as the centre. Wherefore, if from a point, &c. Recite (a) def. 16, 1; (b) p. 7, 3. Q. E. D. 10 Th. One circumference of a circle cannot cut another in more than two points. If it be possible, let the circumferences AВС, DEF mutually intersect in three points, B, G, F; and let K be the centre of the circle ABC, having F KB, KG, KF as equal radii. D C Now, from a point K, within the circle DEF, there are drawn to the circumference more than two equal straight lines: the point K is therefore H 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. Recite (a) p. 9, 3; (6) p. 5, 3. |