Let ABC be a circle, and BEC an angle at the centre, and Book III. BAC an angle at the circumference, which have the same circumference BC for their base; the angle BEC is double of the angle BAC, First, let E the centre of the circle be within the angle BAC, and join AE, and produce it to F: because EA is equal to EB, the angle EAB is equal a to the angle EBA; therefore the angles EAB, EBA are double of the angle EAB; but the angle BEF is equal to the angles EAB, EBA; therefore also the angle BEF is double of the angle EAB: for the same reason, the angle FEC is double of the B F A a 5. 1. E b 32. 1 angle EAC: therefore the whole angle BEC is double of the whole angle BAC. Again, Let E the centre of the circle be without the angle BDC, and join DE, and produce it to G. It may be demonstrated, as in the first case, that the angle GEC is double of the angle GDC, and that GEB a part of the first is double of GDB a part of the other; therefore the remaining angle BEC is double of the remaining angle BDC. Therefore, the angle at the centre, &c. Q. E. D. G A D E B C PROP. XXI. THEOR. THE angles in the same segment of a circle are See N. equal to one another. Book III. the circumference, viz. BCD, for their base; therefore the angle BFB is double a of the angle BAD: for the same reason, a 20.3. the angle BFD is double of the angle BED: therefore the angle BAD is equal to the angle BED. B A E 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 AF 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 D F whole angle BED. Wherefore, the C angles in the same segment, &c. Q. E. D. PROP. XXII. THEOR. 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 AC, BD; and because the three angles of every tria 32. 1. angle are equal to two right angles, the three angles of the triangle CAB, viz. the angles CAB, ABC, BCA are equal to two right angles: but the angle CAB b 21. 3. is equal to the angle CDB, because angle ADB, because they are in the D C whole angle ADC is equal to the A B equal to the angles ABC, ADC: but ABC, CAB, BCA are equal to two right angles; therefore also the angles ABC, ADC are equal to two right angles: in the same manner, the angles BAD, DCB may be shown to be equal to two right angles. Book III. Q. E. D. PROP. XXIII. THEOR. UPON the same straight line, and upon the same See N. side of it, there cannot be two similar segments of circles not coinciding with one another. D a 10.3. If it be possible, let the two similar segments of circles, viz. ACB, ADB, be upon the same side of the same straight line AB, not coinciding with one another: then, because the circle ACB cuts the circle ADB in the two points A, B, they cannot cut one another in any other point 2: one of the segments must therefore fall within the other; let ACB fall within ADB, and draw the straight line BCD, and join CA, DA: and because the segment ACB is similar to the segment ADB, and that similar segments of circles contain equal an- b 11. def. gles; the angle ACB is equal to the angle ADB, the exterior to 3. the interior, which is impossible c. Therefore, there cannot be c 16. 1. two similar segments of a circle upon the same side of the same line, which do not coincide. Q. E. D. A B PROP. XXIV. THEOR. SIMILAR segments of circles upon equal straight See N. lines are equal to one another. Let AEB, CFD be similar segments of circles upon the equal straight lines AB, CD; the segment AEB is equal to the segment CFD. C, and the straight A B C D line AB upon CD, the point B shall coincide with the point D, Book III. because AB is equal to CD: therefore, the straight line AB coinciding with CD, the segment AEB must a coincide with the a 23.3. segment CFD, and therefore is equal to it. Wherefore, similar segments, &c. Q. E. D. PROP. XXV. PROB. See N. a 10. 1. c 6. 1. 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. Bisecta AC in D, and from the point D draw b DB at right b 11. 1. angles to AC, and join AB: first, let the angles ABD, BÅD be equal to one another; then the straight line BD is equal to DA, and therefore to DC; and because the three straight lines DA, DB, DC are all equal, D is the centre of the circled: 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 ABC is a segment is described: and because the centre D is in AC, the segment ABC is a se d 9.3. micircle but if the angles ABD, BAD are not equal to one e 23. 1. another, at the point A, in the straight line AB, make the angle f 4. 1. BAE equal to the angle ABD, and produce BD, if necessary, to E, and join EC: and because the angle ABE is equal to the angle BAE, the straight line BE is equal to EA: and because AD is equal to DC, and DE common to the triangles ADE, 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 AE is equal f to the base EC: but AE was shown to be equal to EB; wherefore also BE is equal to EC: and the three straight lines AE, EB, EC are therefore equal to one another; wherefore d E is Book III. the centre of the circle. From the centre E, at the distance of any of the three AE, EB, EC, describe a circle; this shall pass d 9. 3. through the other points; and the circle of which ABC 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. PROP. XXVI. THEOR. 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 H; therefore the base BC is equal to the base EF: and because the angle at A is equal a 4. 1. to the angle at D, the segment BAC is similar to the segment b 11. def. EDF;d they are upon equal straight lines BC, EF; but simi- 3. lar se goents of circles upon equal straight lines are equal to one amer: therefore the segment BAC is equal to the segment c 24. & EDF: but the whole circle ABC is equal to the whole DEF; M |