T PROP. XXIII. THEOR. HE oppofite angles of every quadrilateral figure infcri- Book III. Let ABDC be a quadrilateral figure infcribed in the circle See fig. 22. ABDC, the oppofite angles BAC, BDC are equal to two right angles; as alfo ABD, ACD; join DA, EC. Then in the fegment DBAC, the angle DBC is equal to DAC; for the fame reafon the angle BAD is equal to a 22. BCD; therefore the whole angle BAC is equal to the two angles DBC, DCB; add BDC to both; then the two angles BAC, BDC are equal to the three angles in the triangle BDC, that is, equal to two right angles b: But all the in-b 32. 1. ward angles of any quadrilateral figure are equal to four right angles; therefore the angles ABD, ACD are equal c Cor. 32. I, to two right angles. Wherefore, &c. T PROP. XXIV. THE OR. WO fimilar and unequal fegments of circles cannot be placed upon the fame right line, either on the Jame or oppofite fides. For, if poffible, let the fimilar fegments ADB, ACB, be see fig. 23, placed upon the fame right line AB; if not on the fame fide, there can be drawn, on the fame fide, a fegment equal to one of them; let this be ACB; then, because they are fimilar, the angle ACB is equal to ADB a; which cannot be b. a Def. II. Wherefore, &c. S PROP. XXV. THE OR. Imilar fegments of circles, being upon equal right lines, are, equal to one another. b 16. I Let AEB, CFD 'be fimilar fegments, conftitute on the See fig. 24, equal right lines AB, CD; they are equal to one another. For, let the fegment AEB be applied to the fegment CFD, fo as the point A coincide with C, and B with D; then AB will coincide with CD, and the fegment AEB with CFD : If not, they will cut one another, which let be in G; then the Book III. the fegment CGD cuts the fegment CFD in the points C, G, D; therefore a circle will cut another circle in more than two points, which cannot be. Wherefore, &c. a 10. a Def. I. b4 5. PROP. XXVI. THE OR. IN equal circles, the circumferences, upon which equal angles ftand, are equal to one another, whether the angles are at the center or circumferences. Let ABC, DEF, be equal circles, and BGC, EHF, equal angles at the centers, and BAC, EDF at the circumferences; then the circumference BKC is equal to ELF; join BC, EF; then, fince BG, GC are equal to EH, HF a, and the angle BGC to EHF, the bafe BC is equal to EF b; and because the angle BAC, is equal to EDF, and the right line BC to EF, the fegment BAC is fimilar, and equal to EDF c; but the and Prop. whole circle BAC is equal to the circle EDF, and the circumference BAC to the circumference EDF; therefore the remainder BKC is equal to ELF. Wherefore, &c. 25. Def. 10. 46. PROP. XXVII. THEOR. Ngles, that ftand upon equal circumferences in equal circles, are equal to each other, whether they be at the centers or circumferences. Let the angles at the centers of the circles ABC, DEF, be BGC, EHF, and the angles BAC, EDF, at their circumferences, ftanding on the equal circumferences BC, EF; then the angle BAC is equal to the angle EDF, and BGC to EHF. For, if the angle BGC be not equal to the angle EHF, let one of them be greater, as BGC, and make BGK equal to EHF; then the circumference BK is equal to EF a; but EF Hypoth. is equal to BC b; therefore, BK is equal to BC, a part to the whole, which is impoffible; therefore, the angle BGK is not equal to EHF; therefore, no angle but BGC can be equal to EHF at the center, and BAC to EDF at the circumference. Wherefore, &c. PRO P. XXVIII. and XXIX. THE OR. N equal circles, equal right lines cut off equal circumferences, the greater equal to the greater, and the leffer to the leffer, and the right lines, in equal circles, which cut off equal circumferences, are equal. Let ABC, DEF be equal circles, in which are the equal right lines BC, EF, which will cut off the greater circumference, viz. BAC equal to EDF, and the leffer BGC to EHF. BOOK III. For, find the centers K and L of the two circles, and join` BK, KC, EL, LF; then, because the two circles are equal, the two fides BK, KC are equal to the two fides EL, LFa, and a Def. 1. the bafe BC equal to EF ; therefore the angle BKC is equal to b Hyp. ELF, and the circumference BGC equal to EHF; but the c 8. i. whole circumference BGCA is equal to the whole circumference EHFD; therefore the remaining circumference, BAC, is is equal to the remaining circumference EDF. Wherefore, &c. And, if the circumference BGC be equal to EHF, the right line BC will be equal to EF; for, the fame conftruction remaining, becaufe BK is equal to KC; and EL to LF, and the angle BKC to ELF, the bafe BC is equal to EF. Wherefore, d 27. &c. C 14. I. To cut PRO P. XXX. PRO B. O cut a given circumference into two equal parts. It is required to cut the given circumference ADB into two equal parts. Join AB, which bifect in C *; from which draw a 10. I. the right line CD at right angles to AB, and join AD, b 11. 1. DB. Now, because AC is equal to CB, and CD common, the two fides AC, CD are equal to the two fides BC, CD, and the angle ACD to BCD; therefore, the bafe AD is equal to DB, and the circumference AD to DB. Wherefore, &c, C 4.1. d 28. BOOK III. a 5. I. THE angle in a femicircle is a right angle, and the angle in a Segment greater than a femicircle is lefs than a right angle, and the angle in a fegment less than a femicircle, is greater than a right angle. Let the angle BAC be an angle in a femicircle, fstanding on the diameter BC; find the center E, and join AE; the angle BAC will be a right angle: Let ADC be a fegment cut off by the right line AC, join AD, DC; then the angle ADC is greater than a right angle; if the circle be compleated, the fegment ABC is greater than a femicircle, and the angle ABC in it lefs than a right angle. For, because E is the center of the circle, the angle ABE is equal to BAE 2, and the angle EAC to ACE; therefore, the whole angle BAC is equal to the two angles ACB, ABC; b Cor. 32. therefore, BAC is a right angle, and the angle BAC is great I. C 22. er than ABC; for BAE is equal to ABC. But the angles ABC, ADC are equal to two right angles, and ABC is lefs than a right angle; therefore ADC is greater than a right angle. Wherefore, &c. Cor. Hence the angle of a fegment greater than a femicircle is greater than a right angle, and the angle of a fegment less than a femicircle, is lefs than a right angle. For the angle that the circumference BA makes with the right line AC is greater than a right angle; for it contains the right angle BAC. And the angle that the circumference AC makes with the right line AC, is less than a right angle; for, if BA be produced to F, the right angle FAC contains it. PROP. XXXII. THEOR. IF a right line touch a circle, and from the point of contact a right line be drawn to the circle, the angles that right line makes with the tangent are equal to the angles in the alternate fegments of the circle. Let the right line EF touch the circle ABCD in the point B; from any point D, in the circle, draw the right line DB; then the angle DBF is equal to the angle in the alternate segment DAB; and the angle DBE equal to DCB; for, from the point of contact B, draw BA at right angles to EF; take any point Book III. C in the circumference, and join AD, DC, CB. a II. I. Now, because BA is drawn from B, at right angles, to EF, the center of the circle is in AB; and, becaufe ADCB is a femicircle, b 19. the angle ADB is a right angle ; therefore ADB is equal to the c 31. two angles DBA, DĂBa; but ABF is likewife a right angle; der. 32. 1. therefore the angle ABF is equal to the angles DBA, DAB; but the angle ABF is likewife equal to the angles DBF, DBA; therefore the angles DBA, DBF, are equal to the angles DAB, DBA. Take the common angle DBA from both, there re- © ax. I. I. mains the angle DBF equal to DAB, the angle in the alternate fegment. Likewife the angle DCB is equal to the angle DBE; for, DCB, DAB, are equal to two right angles f, and DBF, DBE §, f 22. equal to two right angles; but DAB is proved equal to DBF;8 13. 1. therefore the remainder, DCB, is equal to DBE. Wherefore, &c. 4. PRO P. XXXIII. PRO B. PON a given right line to defcribe a fegment of a circle, It is required, upon AB, to describe a fegment of a circle, that will contain an angle equal to a given angle, C. b II. I. C IO. I, At the point A, with the right line AB, make the angl BAD equal to C2; draw AE at right angles to AD; bifecta 23. 1. AB in F, and draw FG, at right angles, to AB, cutting AE in the point G; join GB; with the center G, and distance GA, defcribe the circle ABE, which will pals through the point B; for, because AB is bifected in F, and GF drawn, at right angles, to AB, the right lines AF, FG. are equal to BF, FG; and the angle AFG equal to BFG; therefore AG is equal to GB d. Now, because AD is a tangent to the d 4. 1. circle, the angle BAD is equal to the angle in the alternate e 16. fegment BEA; but the angle DAB is equal to the angle C; f 32. therefore the angle AEB is equal to the angle C. Wherefore, &c. T PRO P. XXXIV. PRO B. O cut off a fegment from a given circle that shall contain an g Conft. Is |