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). B C C E 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 conFC is therefore perpendicular to DE. (b) p. 12, 1; tact. Recite (a) p. 1,3; (d) p. 18, 1; (e) def. 15, 1; D Q. E. D. (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. B 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. Recite (a) p. 18, 3; (c) ax. 10; 20 Th. (b) def. 10, 1; D C Q. E. D. E 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 circumference, and upon the same arcs as the former. Draw the diameter DEG. 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. Q. E. D. 21 Th. 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): therefore BAD is equal to BED. 2. If the centre F be not in the segment, draw the diameter AFC, and join CE: then, the angles BAC, BEC, in the same segment are equal, by case B 1; and the angles CAD, CED are equal, for the same reason: therefore, BAC and CAD are equal to BEC and CED; that is, BAD is equal to BED. Wherefore, the angles in the same segment, &c. Q. E. D. Recite (a) p. 20, 3; (b) ax. 7. AE 22 Th. The opposite angles of any quadrilateral figure inscribed in a circle (ABCD) are together equal to two right angles. In the quadrilateral figure ABCD, the two angles ABC, ADC are equal to two right angles: join AC and BD. In the triangle ABC, the three angles are equal to two right angles (a): but the angles ACB, ADB, as also the angles BAC, BDC are equal (b); there- A fore, the two ACB and BAC are equal to the two BDC and ADB; that is, to ADC: to each side add D B ABC; then the three ACB, BAC, ABC are equal to the two ADC and ABC; but the three are equal to two right angles; and so, the two are equal to the same. In this manner, it may be shown that BAD and BCD are also equal to two right angles. Therefore, the opposite angles, &c. Recite (a) p. 32, 1; (b) p. 21, 3. Q. E. D. 23 Th. Upon one side of the same chord (AB) no two similar segments of circles can be described, which shall not coincide with each other. If possible, let the segments ACB, ADB, which are on the same side of AB, and do not coincide, be similar. Then, since the circles, of which these are segments, can cut each other only in two points (a); n the segments may meet each other only in the points A, B. Draw the straight line BCD; and join AC, AD. B Now similar segments contain equal angles (b); therefore the angles ACB, ADB are equal: but ADC is a triangle, and its exterior angle is ACB, which exceeds the interior ADB, (c): therefore, since one angle cannot be equal to, and greater than another, the segments ACB, ADB are not similar. Therefore, upon one side of the same chord, &c. Q. E. D. 24 Th. Similar segments of circles (AEB, CFD), upon equal chords (AB, CD), are equal to each other. Argument. If the segment AEB be applied to the segment CFD, because the chords are equal (a), their extreme points A, B, and C, D, shall coincide: but the same points are the extremities of the arcs, which must also coincide; because the segments A are similar (b): therefore, the perimeters everywhere coincide and bound the same space (c). Therefore, similar segments, upon equal chords, Q. E. D. &c. Recite (a) def. 3, 1; (c) def. 37, 1. (b) def. 7, and p. 23, 3; DD 25 P. A segment of a circle (ABC) being given, to find the centre, or to describe the circle. 1. Bisect AC in D (a); through D, draw DB at right angles to AC (b); join AB: then if the angles BAD, ABD be equal, the sides DA, DB are equal (c), and D is the centre sought. 2. If DB be less, or greater than DA, or DC, make the angle BAE equal to the angle ABE (d); then EA equals EB (c); join EC. And, beause of the right angles at D; and that DA, DE are equal to DC, DE, the bases EA, EC are equal (e). Therefore, since three equal straight lines, EA, EB, EC, are drawn from the point E to the circumference, E is the centre of the circle (ƒ), of which ABC is a segment. Recite (a) p. 10, 1; (c) p. 6, 1; A B (b) p. 11, 1; (d) p. 23, 1; (e) p. 4, 1; (f) p. 9, 3. 26 Th. In equal circles (ABC, DEF) equal angles at the centres (BGC, EHF), or at the circumferences (BAC, EDF), stand upon equal arcs. Argument. Join BC, EF; then, since the circles are equal, the radii are also equal (a); and since the angles BGC, EHF contained by the equal radii, are equal, the base BC is equal to E the base EF (b). Again, because the angles at A and D are equal, the segments containing them are similar (c); also, because the chords are equal, the segments containing the equal angles are equal (d): therefore, if the equal segments be taken from the equal circles, the segments left will be equal (e):,and so, the segments BKC, ELF are equal: and the arc BKC is equal to the arc ELF. Wherefore, in equal circles, equal angles, &c. Recite (a) def. 15, 1; (d) p. 24, 3; (b) p. 5, 1; (c) ax. 3. Q. E. D. (c) def. 7, 3; 27 Th. In equal circles (ABC, DEF), equal arcs (BKC, ELF) subtend equal angles, whether at the centres (BGC, EHF), or at the circumferences (BAC, EDF). Argument. If the central angles be equal, their halves at the circumferences will be equal (a); but if unequal, one of them, as BGC, is the greater: make BGK equal to EHF (b). Now equal angles stand upon equal arcs (c); therefore, the arc BK CE E is equal to the arc ELF, or its equal BKC: so the part is equal to the whole; but it is not (d). Therefore, BK is not, but BKC is equal to ELF; and the angles BGC, EHF, at the centres, are equal; and their halves, A and D, at the circumferences, are also equal. Wherefore, in equal circles, equal arcs, &c. Q. E. D. Recite (a) p. 20, 3; (c) p. 26, 3; (b) p. 23, 1; (d) ax. 9, 1. 28 Th. In equal circles (ABC, DEF) equal chords (BC, EF) cut off equal arcs; the greater equal to the greater, and the less to the less. Let the greater segments be BAC, EDF, and the less BGC, EHF. Find the centres K, L (a); and draw radii to B, C, and to E, F. Then since the circles are equal, the radii are equal (b); and, since the chords are equal, the angles at K, L, are also equal (c). But B K D equal angles stand upon equal arcs (d); therefore, the arc BGC is equal to the arc EHF. Again, taking the arcs now proved equal, from the equal circumferences, the remaining arcs are equal; namely, BAC to EDF (e). Wherefore, in equal circles, equal chords, &c. Q. E. D. (c) p. 8, 1; Recite (a) p. 1, 3; (d) p. 26, 3; (b) def. 15, 1; |