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 ce: therefore the segment badc 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. : PROPOSITION XXÍI.—THEOREM. The opposite angles of any quadrilateral figure described in a circle are together equal to two right angles. LET abcd be a quadrilateral figure in the circle abcd; any two of its opposite angles are 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 cab, a bc, bca, are equal to two right angles but the angle cab is equal (iii. 21) to the angle c d b, because they are in the same segment bad c, and the angle acb is equal to the angle a d b, because they are in the same segment adcb therefore the whole angle a dc is equal to the angles cab, a cb: to each of these equals b add the angle abc; therefore the angles a bc, cab, bca are equal to the angles a b c, a d c. But abc, cab, bca, are equal to two right angles; therefore also the angles abc, a dc, are equal to two right angles. In the same manner, the angles bad, dcb, may be shewn to be equal to two right angles. Therefore the opposite angles, &c. Q. E. D. a PROPOSITION XXIII. THEOREM. Upon the same straight line and upon the same side of it, there cannot be two similar segments of circles not coinciding with one another. d b Ir it be possible, let the two similar segments of circles, viz. a cb, adb be upon the same side of the same straight line a b, not coinciding with one another. Then because the circle a cb cuts the circle a db 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. Let acb fall within a db, and draw the straight line bed, and join ca, da. And because the segment acb is similar to the segment ad b, and that similar segments of circles contain (iii. def. 11) equal angles; the angle a cb is equal to the angle adb, 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. Similar segments of circles upon equal straight lines are equal to one another. LET aeb, cfd be similar segments of circles upon the equal straight lines ab, cd; the segment a eb is equal to the segment cfd. For if the segment a eb be applied to the segment cfd, so as the point a be on C, and the straight line ab upon cd, the point b shall coincide with the point d, because ab is equal to cd. Therefore the straight line ab coinciding with cd, the segment a eb must (iii. 23) coincide with the segment cfd, Wherefore similar segments, &c. Q. E. D. and therefore is equal to it. PROPOSITION XXV.-PROBLEM. A segment of a circle being given, to describe the circle of which it is the segment. LET abc 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 abc is a semicircle. But if the angles abd, bad are not equal to one another, at the point to ec. a, in the straight line ab, make (i. 23) the angle bae equal to the angle abd, and produce bd, if necessary, to e, and join ec. And because the angle a be is equal to the angle bae, the straight line be is equal (i. 6) to e a And because ad is equal to dc, and de common to the triangles ade, cde, the two sides a d. 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 ae is equal (i. 4) to the base ec: but ae was shewn to be equal to eb, wherefore also be is equal And the three straight lines ae, 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 bc is a segment is described. And it is evident that if the angle abd be greater than the angle bad, the centre e falls without the segment abc, which therefore is less than a semicircle. But if the angle abd be less than bad, the centre e falls within the segment abc, 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. PROPOSITION XXVI.-THEOREM. In equal circles, equal angles stand upon equal circumferences, whether they be at the centres or circumferences. LET abc, 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 abc, 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 a d b g f 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 edf. 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 edf. 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 bc, 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 bgc, 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 abc, def be equal circles, and bc, ef equal straight lines in them, which cut off the two greater circumferences bac, edf, and the two less bgc, ehf; the greater bac is equal to the greater edf, and the less bgc to the less ehf. Take (i. 3) k, l, the centres of the circles, and join bk, k c, el, lf. And because the circles are equal, the straight lines from their centres are equal: therefore bk, kc are equal to el, lf; and the base b c is equal to the base ef; therefore the angle bkc is equal (i. 8) to b F a d 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 abc is equal to the whole edf; 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 straight line ef d b g a h Take (i. 3) k, l, the centres of the circles, and join bk, kc, el, lf. And because the circumference bgc is equal to the circumference ehf, the angle bkc is equal (iii. 27) to the angle elf. f And because the circles abc, def are equal, the straight lines from their centres are equal. equal to el, 1f, 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. Therefore bk, kc are PROPOSITION XXX.-PROBLEM. To bisect a given circumference; that is, to divide it into two equal parts. LET adb be the given circumference; it is required to bisect it. Join ab and bisect (i. 10) it in c; from the point c draw cd at right angles to ab, and join a d, db. The circumference adb is bisected in the point d. Because ac is equal to cb, and cd common to the triangles acd, d bcd, the two sides ac, cd are equal to the two bc, cd; and the angle a cd is equal to the angle bcd, because each of them is a right angle therefore the base ad is equal (i. 4) to the base bd. But equal straight lines cut off equal (iii. 28) circumferences, the greater equal to the greater, and the less to the less, and ad, 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 `d b. Therefore the given circumference is bisected in d. Which was to be done. a b |