b 25.6. c 21. 6. Book VI. ted to D: And make the parallelogram GH equal to EL and C together, and fimilar and fimilarly fituated to D; wherefore GH is fimilar to EL: Let KH be the fide homologous to FL, and KG to FE: And because the parallelogram GH is greater than EL, therefore the fide KH is greater than FL, and KG than FE: Produce FL and FE, and make FLM equal to KH, and FEN to KG, and complete the parallelogram MN. MN is therefore equal and fimilar to GH; but GH is fimilar to EL; wherefore MN is fimilar to K H e 36. I. f 43. 1. 5 24. 6. MN is equal to EL and C: Take away the common part EL; then the remainder, viz. the gnomon NOL, is equal to C. And becaufe AE is equal to EB, the parallelogram AN is equal to the parallelogram NB, that is, to BM. Add NO to each; therefore the whole, viz. the parallelogram AX, is equal to the gnomon NOL. But the gnomon NOL is equal to C; therefore alfo AX is equal to C. Wherefore to the ftraight line AB, there is applied the parallelogram AX equal to the given rectilineal C, exceeding by the parallelogram PO, which is fimilar to D, because PO is fimilar to EL. Which was to be done. PROP. XXX. P R O B. To cut a given straight line in extreme and mean ratio. Let AB be the given straight line; it is required to cut it in extreme and mean ratio. Upon Book VI. m a 46. 1. b 29. 6. D E B C 14. 6. d 34. I. Upon AB defcribe the fquare BC, and to AC apply the parallelogram CD equal to BC, exceeding by the figure AD fimilar to BC: But BC is a fquare, therefore alfo AD is a fquare; and becaufe BC is equal to CD, by taking the common part CE from each, the remainder BF is equal to the remainder AD: And thefe figures are equiangu"lar, therefore their fides about the equal angles are reciprocally proportional: Wherefore as FE to ED, fo AE to EB: But FE is equal to AC d, that is, to AB; and ED is equal to AE: Therefore as BA to AE, fo is AE to EB: But AB is greater than AE; wherefore AE is greater than EB: Therefore the straight line AB is cut in ex- 14. 5. treme and mean ratio in Ef. which was to be done. Otherwise, C F Let AB be the given ftraight line; it is required to cut it in extreme and mean ratio. Divide AB in the point C, fo that the rectangle contained by AB, BC be equal to the fquare of AC3. Then, because the rectangle AB, BC is equal A to the fquare of AC, as BA to AC, fo is f 3. def. 6. C B AC to CB: Therefore AB is cut in extreme and mean ratio h 17. 6. in Cf. Which was to be done. IN right angled triangles, the rectilineal figure defcribed See N. upon the fide oppofite to the right angle, is equal to the fimilar, and fimilarly defcribed figures upon the fides containing the right angle. Let ABC be a right angled triangle, having the right angle BAC: The rectilineal figure defcribed upon BC is equal to the fimilar, and fimilarly defcribed figures upon BA, AC. Draw the perpendicular AD; therefore, because in the right angled triangle ABC, AD is drawn from the right angle at A perpendicular to the bafe BC, the triangles ABĎ, ADC are fimilar to the whole triangle ABC, and to one another3, a 8. 6. and b 4.6. Book VI. and because the triangle ABC is fimilar to ABD, as CB to BA, fo is BA to BD; and because these three ftraight lines are proportionals, as the first to the third, fo is the figure upon the first to the fimilar, and fimilarly defcribed figure upon c 2. Cor. the fecond: Therefore as CB to BD, fo is the figure upon as BD and DC together to BC, fo are the figures upon BA, AC to that upon BC: But BD and DC together are equal to BC. Therefore the figure defcribed on BC is equal f to the fimilar and fimilarly defcribed figures on BA, AC. Wherefore, in right angled triangles, &c. Q. E. D. See N. a 29. I. IF two triangles which have two fides of the one proportional to two fides of the other, be joined at one angle, fo as to have their homologous fides parallel to one another; the remaining fides fhall be in a ftraight line. Let ABC, DCE be two triangles which have the two fides BA, AC proportional to the two CD, DE, viz. BA to AC, as CD to DE; and let AB be parallel to DC, and AC to DE. BC and CE are in a straight line. Because AB is parallel to DC, and the straight line A the triangles ABC, DCE have one angle at A equal to one at Book VI. D, and the fides about thefe angles proportionals, viz. BA to AC, as CD to DE, the triangle ABC is equiangular to DCE: b 6. 6. Therefore the angle ABC is equal to the angle DCE: And the angle BAC was proved to be equal to ACD: Therefore the whole angle ACE is equal to the two angles ABC, BAC; add the common angle ACB, then the angles ACE, ACB are equal to the angles ABC, BAC, ACB: But ABC, BAC, ACB are equal to two right angles; therefore alfo the angles ACE, c 32. 2. ACB are equal to two right angles: And fince at the point C, in the ftraight line AC, the two ftraight lines BC, CE, which are on the oppofite fides of it, make the adjacent angles ACE, ACB equal to two right angles; therefore d BC and d 14. 1 CE are in a ftraight line. Wherefore, if two triangles, &c. Q. E. D. PROP. XXXIII. THEOR. IN equal circles, angles, whether at the centers or cir- see N. cumferences, have the fame ratio which the circumferences on which they ftand have to one another: So alfo have the fectors. Let ABC, DEF be equal circles; and at their centers the angles BGC, EHF, and the angles BAC, EDF at their circumferences; as the circumference BC to the circumference EF, fo is the angle BGC to the angle EHF, and the angle BAC to the angle EDF; and alfo the fector BGC to the fector EHF. Take any number of circumferences CK, KL, each equal to BC, and any number whatever FM, MN each equal to EF; And join GK, GL, HM, HN. Because the circumferences BC, CK, KL are all equal, the angles BGC, CGK, KGL are alfo all equal: Therefore what multiple foever the circum- a 27. Ji ference BL is of the circumference BC, the fame multiple is the angle BGL of the angle BGC: For the fame reason, whatever multiple the circumference EN is of the circumference EF, the fame multiple is the angle EHN of the angle EHF: And Book VI. a 27. 3. And if the circumference BL be equal to the circumference EN, the angle BGL is alfo equal to the angle EHN; and if the circumference BL be greater than EN, likewife the angle BGL is greater than EHN; and if lefs, lefs: There being then four magnitudes, the two circumferences BC, EF, and the two angles BGC, EHF; of the circumference BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the circumference BL, and the angle BGL; and of the circumference EF, and of the angle EHF, any equimultiples what ever, viz. the circumference EN, and the angle EHN: And it has been proved, that, if the circumference BL be greater than EN, the angle BGL is greater than EHN; and if equal, equal; and if lefs, lefs: As therefore the circumference 5. Def, s. BC to the circumference EF, fob is the angle BGC to the angle EHF: But as the angle BGC is to the angle EHF, fo is the angle BAC to the angle EDF, for each is double of each: Therefore, as the circumference BC is to EF, fo is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. C 15. 6. d 20. 3. € 4. I, Alfo, as the circumference BC to EF, fo is the sector BGC to the fector EHF. Join BC, CK, and in the circumferences BC, CK take any points X, O, and join BX, XC, CO, OK : Then, because in the triangles GBC, GCK the two fides BG, GC are equal to the two CG, GK, and that they contain equal angles; the bafe BC is equal to the bafe CK, and the triangle GBC to the triangle GCK: And because the circumference BC is equal to the circumference CK, the remaining part of the whole circumference of the circle ABC, is equal to the remaining part of the whole circumference of the fame circle: Wherefore the angle BXC is equal to the angle COK'; f 11. def. 3. and the fegment BXC is therefore fimilar to the fegment COK; and |