But if the segment baed be not greater than a semicircle, let bad, bed be angles in it; these also are equal to one another : draw a f to the centre, and produce it to c, and join če : therefore the segment bad c is greater than a semicircle ; and the angles in it bac, bec are equal, by the first case : for the same reason, because cbed is greater than a semicircle, the angles cad, ced, are equal : therefore the whole angle bad is equal to the whole angle bed. Wherefore the angles in the same segment, &c. Q. E. D. 1 PROPOSITION XXII. -THEOREM. together equal to two right angles. Join a c, bd; and because the three angles of every triangle are equal (i. 32) to two right angles, the three angles of the triangle ca b, viz. the angles ca b, a b c, b ca, are equal to two right But a b c, ca b, bca, are equal to two right angles ; therefore also the angles a b c, ad c, are equal to two right angles. In the same manner, the angles b a d, dob, may be shewn to be equal to two right angles. Therefore the opposite angles, &c. Q. E. D. PROPOSITION XXIII.—THEOREM. IF it be possible, let the two similar segments of the same side of the same straight line a b, not coinciding with one another. Then because the circle a cb cuts the circle adb in the two points a, b they cannot cut one another in any other point (iii. 10). One of the segments must therefore fall within the other. upon Let a cb fall within a db, and draw the straight line bed, and join ca, da. And because the segment acb is similar to the segment a db, and that similar segments of circles contain (iii. def. 11) equal angles ; the angle acb is equal to the angle a db, the exterior to the interior, which is impossible (i. 16). Therefore there cannot be two similar segments of a circle upon the same side of the same line, which do not coincide. Q. E. D. PROPOSITION XXIV.—THEOREM. ariother. LET à eb, cfd be similar segments of circles upon the equal straight lines a b, cd; the segment a eb is equal to the segment ofd. For if the segment a eb f e be applied to the segment cfd, so as the point a be on C, and the straight line a b upon cd, the point b shall d b coincide with the point d, because a b is equal to cd. Therefore the straight line a b coinciding with cd, the segment a eb must (iii. 23) coincide with the segment cfd, and therefore is equal to it. Wherefore similar segments, &c. Q. E. D. a PROPOSITION XXV.-PROBLEM. A segment of a circle being given, to describe the circle of which it is the segment. LET a b c be the given segment of a circle; it is required to describe the circle of which it is the segment. Bisect (i. 10) ac in d, and from the point d draw (i. 11) db at right angles to a C, and join a b. First, let the angles a bd, bad be equal to one another; then the straight line bd is equal (i. 6) to da, and therefore to dc; and because the three straight lines da, db, dc, are all equal, d is the centre of the circle (iii. 9). From the centre d, at the distance of any of the three da, db, dc, describe a circle; this shall pass through the other points; and the circle of which a bc is a segment is described. And because the centre d is in a c, the segment a bc is a semicircle. b d e d à d But if the angles a bd, bad are not equal to one another, at the point to e ao to ec. a, in the straight line ab, make (i. 23) the angle ba e equal to the angle a bd, and produce bd, if necessary, to e, and join ec. And because the , angle a be is equal to the angle ba e, the straight line be is equal (i. 6) And because a d is equal to dc, and de common to the triangles a de, cde, the two sides ad, de are equal to the two cd, de, each to each ; and the angle ade is equal to the angle cde, for each of them is a right angle; therefore the base a e is equal (i. 4) to the base ec: but a e was shewn to be equal to e b, wherefore also be is equal And the three straight lines a e, eb, ec are therefore equal to one another; wherefore (iii. 9) e is the centre of the circle. From the centre e, at the distance of any of the three a e, eb, e c, describe a circle, this shall pass through the other points; and the circle of which a b c is a segment is described. And it is evident that if the angle a bd be greater than the angle bad, the centre e falls without the segment a b c, which therefore is less than a semicircle. But if the angle a bd be less than bad, the centre e falls within the segment a b c, which is therefore greater than a semicircle. Wherefore, a segment of a circle being given, the circle is described of which it is a segment. Which was to be done. ز PROPOSITION XXVI.—THEOREM. In equal circles, equal angles stand upon equal circumferences, whether they be at the centres or circumferences. LET a b c, def be equal circles, and the equal angles bgc, ehf at their centres, and bac, edf at their circumferences. The circumference bkc is equal to the circumference elf. Join bc, ef; and because the circles a b c, def, are equal, the straight lines drawn from their centres are equal. Therefore the two sides bg, gc, are equal to the two eh, hf, and the angle at g is equal to the angle at а. d ong b f 1 h; therefore the base bc is equal (i. 4) to the base ef. And because the angle at a is equal to the angle at d, the segment bac is similar (iii . def. 11) to the segment edf; and they are upon equal straight lines bc, ef. But similar segments of circles upon equal straight lines are equal (iii. 24) to one another; therefore the segment bac is equal to the segment ed f. But the whole circle abc is equal to the whole def; therefore the remaining segment bkc is equal to the remaining segment elf, and the circumference bkc to the circumference elf. Wherefore, in equal circles, &c. Q. E. D. PROPOSITION XXVII.—THEOREM. In equal circles, the angles which stand upon equal circumferences are equal to one another, whether they be at the centres or circumferences. LET the angles bgc, ehf at the centres, and bac, edf at the circumferences of the equal circles abc, def, stand upon the equal circumferences bc, ef. The angle bgc is equal to the angle ehf, and the angle bac to the angle ed f. If the angle bgc be equal to the angle ehf, it is manifest (iii. 20) that the angle bac is also equal to edf. But if not, one of them is the greater. Let bgc be the greater, and at the point g, in the straight line bg, make (i. 23) the angle bgk equal to the angle ehf; but equal angles stand upon equal circumferences (iii. 26), when they are at the centre; therefore the circumference bk is equal to the circumference ef. But ef is equal to bc; therefore also bk is equal to be, the less to the greater, which is impossible. Therefore the angle bgc is not unequal to the angle ehf; that is, it is equal to it. And the angle at a is half of the angle bg.c, and the angle at d half of the angle ehf: therefore the angle at a is equal to the angle at d. Wherefore, in equal circles, &c.Q. E. D. PROPOSITION XXVIII.-THEOREM. In equal circles, equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less. LET a b c d e f be equal circles, and bc, ef equal straight lines in them, which cut off the two greater circumferences bac, ed f, and the two less bgc, ehf; the greater bac is equal to the greater edf, and the less bgc to the less ehf. a d Take (i. 3) k, 1, the centres of the circles, and join bk, k c, el, lf. And because the circles are equal, the straight lines k 1 from their centres are equal: therefore bk, kc are equal to b el, lf; and the base b c is equal to the base ef; therefore the angle bkc is equal (i. 8) to 6o h F the angle elf. But equal angles stand upon equal (iii. 26) circumferences, when they are at the centres; therefore the circumference bgc is equal to the circumference ehf. But the whole circle a bc is equal to the whole ed f; the remaining part therefore of the circumference, viz. bac, is equal to the remaining part edf. Therefore, in equal circles, &c. Q. E. D. PROPOSITION XXIX.-THEOREM. In equal circles equal circumferences are subtended by equal straight lines. LET abc, def be equal circles, and let the circumferences bgc ehf also be equal ; and join bc, ef. The straight line bc is equal to the a straight line e f. d Take (i. 3) k, 1, the centres of the circles, and join bk, kc, el, lf. And because the circumk 1 ference bgc is equal to the cir cumference ehf, the angle bkc b ъ is equal (iii. 27) to the angle elf. f And because the circles abc, 8 def are equal, the straight lines h from their centres are equal. Therefore bk, kc are equal to el, lf, and they contain equal angles : therefore the base bc is equal (i. 4) to the base ef. Therefore, in equal circles, &c. Q. E. D. PROPOSITION XXX.-PROBLEM. To bisect a given circumference; that is, to divide it into two equal parts. LET a db be the given circumference; it is required to bisect it. Join a b and bisect (i. 10) it in c; from the point c draw cd at right angles to a b, and join a d, db. The circumference a db is bisected in the point d. Because ac is equal to cb, and cd common to the triangles a cd, bcd, the two sides a c, cd are equal to the two d bc, cd; and the angle a cd is equal to the angle bcd, because each of them is a right angle: therefore the base a d is equal (i. 4) to the base bd. But equal straight lines cut off equal (iii. 28) a C b circumferences, the greater equal to the greater, and the less to the less, and a d, db are each of them less than a semicircle ; because cd passes through the centre (i. 3. cor.). Wherefore the circumference a d is equal to the circumference db. Therefore the given circumference is bisected in d. Which was to be done. |