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. 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. D C 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 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 B 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. Q. E. D. Recite (a) p. 32, 1; (b) p. 21, 3. 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); the segments may meet each other only in the B points A, B. Draw the straight line BCD; and join AC, AD. 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 (6): therefore, the perimeters everywhere coincide and bound the same space (c). Therefore, similar segments, upon equal chords, &c. Recite (a) def. 3, 1; Q. E. D. E B F (b) def. 7, and p. 23, 3; (c) def. 37, 1. 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 (f), of which ABC is a segment. Recite (a) p. 10, 1; (b) p. 11, 1; (c) p. 6, 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. Again, because the angles at A K 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 Wherefore, in equal circles, equal angles, &c. ELF. 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). arcs (c); therefore, the are BK 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; (b) p. 23, 1; (c) p. 26, 3; (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. at K, L, are also equal (c). But 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 cir Wherefore, in equal circles, equal chords, &c. cumferences, the remaining arcs are equal; namely, BAC to EDF (e). Q. E. D. Recite (a) p. 1, 3; (b) def. 15, 1; (c) p. 8, 1; (d) p. 26, 3; (e) ax. 3, 1. 29 Th. In equal circles (ABC, DEF), equal arcs are subtended by equal chords. since the triangles BKC, ELF have two sides, and the contained angle in the one equal to the same in the other, their bases are equal (c); namely, the chords BC, EF. Wherefore, in equal circles, equal arcs, &c. Q. E. D. Recite (a) p. 27, 3; (b) def. 15, 1; (c) p. 4, 1. 30 P. To bisect a given arc (ADB); that is, to divide it into two equal parts. Constr. Draw the chord AB, and bisect it in C (a); draw CD at right angles to AB (b); join DA, DB. Argument. Because of the right angles at C, and that CA, CD are equal to CB, CD, the triangles CAD, CBD have their bases equal (c); namely, A the chords AD and BD. Now these equal chords cut off equal arcs (d), less than semicircles; because DC produced is a diameter (e): therefore the given arc is bisected in D, as required. Recite (a) p. 10, 1; (d) p. 28, 3; (e) p. 15, 3. 31 Th. In a circle, the angle in a semicircle is a right angle, in a greater segment the angle is acute; in a less segment the angle is obtuse. Let ABCD be a circle, BC a diameter, E the centre, BAC a semicircle, ABC the greater segment, and ADC the less. Join AE, and produce BA to F. A F D C E 1. The radii EA, EB, EC are opposite to equal angles (a); that is, the angles EAB, EBA, EAC, B ECA are equal to each other; and they are equal two and two: therefore EBA and ECA are equal to EAB and EAC, or the whole angle BAC. But if one angle of a triangle be equal to the other two, it must be a right angle (b); therefore BAC, an angle in a semicircle, is a right angle, as stated. 2. Again, since in the triangle ABC, the two angles EBA, ECA are equal to the right angle BAC, each of them is less than a right an gle: therefore ABC, the angle in the greater segment, is less than a right angle. 3. Also, the quadrilateral ABCD, being inscribed in a circle, any two of its opposite angles are equal to two right angles (c); therefore, since ABC proves to be acute, its opposite ADC is obtuse, and it is in the less segment. Wherefore, in a circle, the angle, &c. Recite (a) p. 5, 1; (b) p. 32, 1; (c) p. 22, 3. Q. E. D. Cor. If an angle of a triangle be equal to its adjacent angle, equal to the other two (p. 32, 1), and is therefore a right angle. it is 32 Th. If a straight line (EF) touch a circle (ABC), and from (B), the point of contact, any chord be drawn in the circle, the angles thus made, on either side, shall be equal to the angles in the alternate segment. 1. If the chord BA pass through the centre (a), each of the segments will be a semicircle, and shall contain a right angle, equal to ABE, or ABF (b). D C 2. If the chord BD make oblique angles with EF (c); join AD: then BDA is an angle in a semicircle, and therefore a right angle, equal to ABF, as before; or to the sum of BAD and ABD (d): from each of these equals take ABD, the remainders DBF and DAB are equal (e); and the latter is in the alternate segment made by DB. E B F 3. At any point C, in the less segment, make an angle BCD; then, since ABCD is a quadrilateral inscribed in a circle, its opposite angles are equal to two right angles (f); therefore BAD, BCD are together equal to DBF, DBE (g): but DBF proves equal to DAB; therefore DBE is equal to BCD, which is in the alternate segment. Wherefore, if a straight line touch, &c. Recite (a) p. 19, 3; (b) p. 31, 3; (d) cor. 31, 3; (e) ax. 3, 1; Q. E. D. (c) Note def. 8, 1; 33 P. Upon a given straight line (AB), to describe a segment of a circle, containing an angle equal to a given rectilineal angle (C). In every case, whether the given angle C be right, or oblique, bisect the given line AB in the point F (a). 1. Let C be a right angle: then, upon F as centre, describe the semicircle AHB; join H to A and B. The angle at H, being in a semicircle, is a right angle (6), and therefore equal to C (c). H A F B |