29 Th. In equal circles (ABC, DEF), equal arcs are subtended by equal chords. Let the arcs BGC, EHF be equal; the chords BC, EF shall also be equal; join K to B and C, and L to E and F. D Then, because the arcs are equal, the central angles are equal (a); and because the circles are equal, the radii are equal (b); and, 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. Recite (a) p. 27, 3; (b) def. 15, 1; (c) p. 4, 1. 30 P. Q. E. D. 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. A B 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, 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; (b) p. 11, 1; (e) p. 15, 3. (r) ax. 10 and p. 4, 1; 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. E F 1. The radii EA, EB, EC are opposite to equal angles (a); that is, the angles EAB, EBA, EAC, 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. if one angle of a triangle be equal to the other two, it must be a right angle (6); therefore BAC, an angle in a semicircle, is a right angle, as stated. But 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, it is equal to the other two (p. 32, 1), and is therefore a right angle. 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). 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 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 (ƒ); 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. Q. E. D. (b) p. 31, 3; (c) Note def. 8, 1; (f) p. 22, 3; Recite (a) p. 19, 3; (d) cor. 31, 3; (e) ax. 3, 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). 2. Let C be any oblique angle (d): make, at the point A, angles BAD, equal to C, and DAE a right angle (e); draw FG at right angles to AB (ƒ); join GB. Now, on account of the right angles at F, and that FA, FG are equal to FB, FG, the bases GA, GB are equal (g); and about G as centre, a circle may be drawn to pass through the points A, E, B, of the greater segment, and A, H, B, of the less; and DA shall touch the circle in A (h). Therefore the angle AEB is equal to BAD, or C, acute; and the angle AHB is equal to BAD, or C, obtuse. Therefore, upon the given straight line, a segment has been described, containing an angle given in variety of magnitude, as right, acute, and obtuse. Recite (a) p. 10, 1; (b) p. 31, 3; (d) Note def. 8,1; (e) p. 23, 1; (g) p. 4, 1, and cor. p. 1, 3; (c) ax. 10, 1; (f) p. 11, 1: (h) p. 31, 32, and cor. p. 16, 3. 34 P. To cut off a segment from a given circle (ABC), which shall contain an angle equal to a given rectilineal angle (D). Constr. Draw any tangent, as EF, touching the given circle in a point B (a); and at the point B, in the straight line FB, make the angle FBC equal to the angle D (b). B E Argument. Because EF touches the circle; and from B, the point of contact, a chord BC is drawn in the circle; there are two angles equal to the two in the alternate segments (c): therefore the angles FBC and BAC are equal; but FBC was made equal to the given angle D; and so, BAC is equal to D (d); and it is in a segment cut off from a given circle; which was to be done. 35 Th. If two chords (AC, BD) in a circle cut each other, the rectangle of the segments (AE, EC) of one of the chords, is equal to the rectangle of the segments (BC, CD) of the other. 1. If the sectional point E, be the centre of the circle; then it is obvious that AEXEC equals BEX ED: for the segments are equal radii (a). 2. If the chords cut each other at right angles in E, but not in the centre, one of them AČ is bisected in E (b), the other passes through the centre F, where it is bisected (c): join AF. Now, since BD is divided equally in F, and unequally in E, BF2, or AF2 BEX EDEF2 (d)=AE2+ EF2 (e): from each of these equals take EF2; then BEXED=AE2, AEXEC: for AE equals EC. 3. If BD pass through the centre F, and cut AC obliquely in E, draw FG perpendicular to AC (b); join AF: therefore, since BD is bisected in F, and AC in G, and both unequally divided in E, BF2=BEXED+EF2(d); and EF2=EG2+GF2(e); AG2=AEXEC+EG2(d); to these add GF2 (ƒ); then, AG+GF2=AEXEC+EG2+GF2=AF2(e)= BF2: therefore, BEXED+EG2+GF2-AEXEC+ EG2+GF2; from which equals taking EG2+GF2, which are common, BEXED=AE×ĚC (ƒ). 4. If neither AC nor BD pass through the centre F, draw the diameter GEFH; then since either of the rectangles AEXEC, or BEXED proves equal A to GEXEH, as above; in this case also AEXEC= BEXED (ƒ). Wherefore, if two chords in a circle, &c. Recite (a) def. 15, 1; (d) p. 5, 2; A Q. E. D. B (b) p. 3, and cor. p. 1,3; (c) def. 14, 1; 36 Th. If from any point (D) without a circle (ABC) two straight lines be drawn, one of which (DCA) cuts, and the other (DB) touches the circle, the rectangle of the secant and its exterior part is equal to the square of the tangent. 1. If the secant DCA, pass through the centre of the circle E, join EB: then EBD is a right angle (a); and DE DB2+BE2 (b); and because AC is bisected in B E and produced to D, DE2=AD×DC+CE2 (c); therefore DB2+BE AD× DC+CE2 (d): take now from both sides the squares of the radii BE, CE, the remainders are equal (e); namely, DB2=ADXDC. 2. If the secant DCA pass on one side of the centre E; join EB, EC, ED, and draw EF perpendicular to AD (f); therefore DE-DF2+FE2 (b)=DB2+BE2, as above and because AC is bisected in Fand produced to D, DF2=AD+DC CF2 (c); to both sides add FE2; therefore DF2+FE2=ADXDC+CF2+FE2 (g): but CF+FECE2 (b), or BE2; substituting the latter, DF2FE2=ADXDC+BE; therefore DB2+BE2= ADXDC+BE2: take from these BE', which is common, DB2=ADXDC (e). Cor. If from any point A, without a circle, two secants AEB, AFC be drawn; the rectangle of the one AB, and its exterior part AE, is equal to the rectangle of the other AC, and its exterior part AF: for each of these rectangles is equal to the square of the tangent AD (d). Wherefore, if from any point without, &c. B Q. E. D. B Recite (a) p. 18, 3; (d) ax. 1; (b) p. 47, 1; (c) p. 6, 2; (ƒ) p. 12, 1; also p. 3, 3; (g) ax. 2. 37 Th. If from a point (D), without a circle (ABC), two straight lines be drawn, one (DB) meeting, and the other (DEA) cutting the circle; if the rectangle of the cutting line and its exterior part (DC), be equal to the square of the line which meets the circle, the latter line shall touch and not cut the circle. Draw the tangent DE (a); find the centre of the circle F (b); join FE, FD, FB: then FED is a right angle (c): and because DE touches and DCA cuts the circle, DE ADX DC (d); but DB2 is given equal to the same rectangle; therefore DB2 equals DE2, and B DE is equal to DB; the radii FB, FE are also equal, and FD is common to the two triangles DBF, DEF; which have therefore three sides in the one equal to Α three sides in the other; therefore (e) the angle DBF is equal to the angle DEF; and the latter being a right angle by construction, the former is also a right angle; and so, the straight line DB touches the circle (f). Therefore, if from a point without a circle, &c. Q. E. D. Recite (a) def. 2, 3; (d) p. 36 3; (b) p. 1, 3; (c) p. 18, 3; (f) p. 16, 3 and cor. |