Book VI, PRO P. X. PROB. To divide a given undivided right line, as another right line is divided. Let the given undivided right line be AB, and the divided line AC, it is required to divide AB as AC is divided. Let AC be any how divided in the points D, E; and making any angle with AB; join BC; and through the points D, E, draw DF, EG, parallel to BC; and through D draw DHK a 31. I. parallel to AB. Then, because FH, HB, are parallelograms, their oppofite fides are equal; and, because FD is parallel to GE, AF is to b 34. 1. FG as AD is to DE . Again, because HE is parallel to BC, c 2. DH is to HK as DE is to EC ; but DH is equal to FG; and and HK to GB; therefore DE is to EC as FG is to GB; but AD is to DE as AF is to FG; wherefore the given undivided line AB is cut in the fame proportion as AC: Which was required. Two PRO P. XI. PROB. WO right lines being given, to find a third proportional. Let AB, AC, be two given right lines, making any angle with each other, it is required to find a third proportional to them. Produce AB, AC, to the points D, E; make BD equal to AC; join the points B, C; through D draw DE parallel to a 3. 1. BC; then AB is to BD as AC is to CE ; but BD is equal to 31.1. AC; therefore AB is to AC as AC is to CE. Wherefore, &c. 2. PRO P. XII. PRO B. T HREE right lines given, to find a fourth proportional. Let A, B, C, be the three given right lines, it is required to find a fourth proportional to them. Let Let DE, EF, be two right lines, making any angle EDF with each other; make DG equal to A; GE equal to B; and DH equal to C. Join GH; and through E draw EF parallel to GH; then DG is to GE as DH is to HF; therefore HF is the fourth proportional required. PRO P. XIII. PROB. To find a mean proportional to two given right lines. Let the two given right lines AB, BC, be placed in one right line, as AC; upon which defcribe a femicircle ADC; at the point B, draw BD at right angles to AC1; join AD, DC; then BD is the mean proportional required. For, because the angle ADC is a right angle, AB is to BD as BD is to BC; therefore BD is a mean proportional to the given right lines AB, BC: Which was to be done. E PRO P. XIV. and XV. THEO R. QUAL parallelograms and triangles, having the one equal to one angle of the other, have the fides about the equal angles reciprocally proportional; and these parallelograms and triangles that have, one angle of the one equal to one angle of the other, and the fides about the equal angles reciprocally proportional, are equal, viz. the parallelogram to the parallelogram, and triangle to the triangle. Let AB, BC, be equal parallelograms, and FBD, EBG, equal triangles, having the angles at B equal; and let the fides DB, BE, be in one right line, and FB, BG, in another; then the tides DB, BE, and GB, BF, that are about the equal angles at B, are reciprocally proportional, that is, DB is to BE as GB to BF. b. Let the equal parallelograms be AB, BC, and compleat the parallelogram EF; then, as the parallelogram AB is to EF, fo is BC to FE; but, as AB is to FE, fo is the base DB to BE ; and, as BC is to FE, fo is GB to BF2; therefore DB is to BE as BC is to FE c. And, if DB is to BE as BG is to BF; then the parallelogram AB is equal to BC; for, as DB is to BE, fo is AB to FE b; and, as GB is to BF, fo is BC to FE; therefore AB is to FE as BC is to FE; therefore AB is equal to BC o. Secondly, let FD, EG, be joined ; then FDB, EBG, are the e- Book VI. qual triangles, and FBE is any third magnitude; therefore FDB is to FBE as EBG is to FBE; but the triangle FDB is to FBE as DB is to BE; and EBG is to FBE as GB is to BFb; b 1. therefore DB is to BE as GB is to BF C. C 11. 5: And, if DB is to BE as GB is to BF, the triangle FDB is equal to EBG: For, as DB is to BE, fo is the triangle FDB to FBE; and, as GB is to BF, fo is the triangle EBG to FBE; therefore FDB is to FBE as EBG is to FBE; therefore the triangle FDB is equal to EBGd: Wherefore equal parallelo- d 9. 5. grams and triangles, &c. PRO P. XVI. THE OR. IF four right lines are proportional, the rectangle contained under the extremes is equal to the rectangle under the means ; and, if the rectangle contained under the extremes be equal to the rectangle contained under the means, then the four right lines are proportional. Let the four right lines AB, CD, E, F, be proportional, fo that AB be to CD as E is to F; then the rectangle under AB, F, is equal to the rectangle under CD, E: For, draw AG equal to F, and at right angles to AB, and CH equal to E, a 11. I. and at right angles to CD; and compleat the rectangles GB, HD: Then, because AB is to CD as CH is to AG, the rec-b 7. 5. tangle BG is equal to HD, and if GB is equal to HD, AB is c 14. to ČD as CH is to AG, that is, as E to F; for the angles at C, A, are equal, being each right ones. Wherefore, &c. PRO P. XVII. THE OR. IF Let the three right lines A, B, C, be proportional, viz. as A is to B fo is B to C; then the rectangle under A, C, is equal to the fquare of B: For, make D equal to B, and compleat the rectangles under A, C, and B, D; then, because AC, BD, are two rectangles, and A is to B as D is to C, AC is a equal to BD; and, because AC is equal to BD, A is to B as M D 7. 5. J. S. b 14. Book VI.D is to C; but B is equal to D; therefore the rectangle under A, C, is equal to the fquare of B. Wherefore, &c. a 23. f. C 4. U PCN a given right line to defcribe a right lined figure fimilar and fimilarly fituated to a right lined figure given. Let AB be the given right line, and CDEFG the right lined figure given; it is required upon AB to defcribe a figure fimilar and fimilarly fituated to CDEFG. Join DG, DF; and, at the points A, B, of the right line AB, make the angles BAH, ABH, equal to the angles C and CDG, each to each; then bcor 32.1 the remaining angles AB, CGD, will be equal b, and the fides about the equal angle proportional, that is, AB to BH as CD to DG; and AH to HB as CG to GD. Again, at the points H, B, with the right line BH, make the angles BHK, HBK, equal to DGF, GDF, each to each; then the remaining third angles HKB, GFD, are equal, and the triangles HKB, FDG, equiangular, and the fides about the equal angles proportional. Again, make the angles BKL, KBL, equal to the angles DFE, FDE, each to each; then the remaining third angles at L, E, will be equal, and the fides about the equal angles proportional ; but all the triangles in the figure ABLKH are proved fimilar to all the triangles in the figure CDEFG; and, because the angles AHB, BHK, are proved equal to the two angles CGD. IGF, each to each, the whole angle AHK is equal to the angle CGF, and the fides about the equal angles proportional; for AH is to HB as CG to GD; and KH to HB as FG to GD; therefore, by equality, AH is to HK as CG is to GF. For the fame reason, HK is to KL as GF to FE; and KL to LB as FE is to ED; therefore the figure ABLKH is fimilar to CDEFG 4. Wherefore, &c. d def. I. PRO P. XIX. THE OR. IMILAR triangles are to one another in the duplicate ratio of their homologous fides. Let ABC, DEF, be fimilar triangles having the angles at B and E equal; and AB, to BC, as DE to EF, and BC the fide homologous to EF; then the triangle ABC to the triangle DEF has a duplicate ratio that BC has to EF. For, take BG a third proportional to BC, EF, that is, BC Book VI. to EF as EF to BG. Join AG; then, because AB is to BC as DE to EF, alter. as AB is to DE fo is BC to EF; but BC a 11. is to EF as EF is to BG; therefore AB is to DE as EF is to BG ; that is, the fides about the equal angles B, E, of the tri- b 11. 5. angles DEF, ABG, are reciprocally proportional; therefore equal to one another; and, because BC is to EF as EF is to c 14. BG, BC has to BG a duplicate ratio of what it has to EF d; d def. 10. 5. and, as BC is to BG fo is the triangle ABC to the triangle ABG; therefore the triangle ABC has to the triangle ABG ae 1. duplicate ratio of what BC has to EF; but the triangle ABG is equal to DEF; therefore ABC is to DEF in the duplicate ratio of BC to EF. Wherefore fimilar triangles, &c. COR. Hence, if three right lines be proportional, as the firft is to the third, fo is a triangle defcribed on the firft, to a fimilar one defcribed on the second. SIM PRO P. XX. THE OR. IMILAR polygons can be divided into an equal number of fimilar triangles, each homologous to the whole; and polygon is to polygon in the duplicate ratio of one homologous fide to the other. Let ABCDE, FGHKL, be fimilar polygons, and AB, FG, two homologous fides; join BE, EC, GL, LH; then the number of triangles in the polygon ABCDE, are equal to the number of triangles in the polygon FGHKL, fimilar to one another, and homologous to the whole ; and the polygon ABCDE will be to the polygon FGHKL in the duplicate ratio of the fide AB to FG. b 6. For, because the polygon ABCDE is fimilar to FGHKL, the angle BAE is equal to GFL; and BA is to AE as GF to FL; and the angle ABE equal to FGLb; and AB to BE as a 4 FG to GL; but the whole angle ABC is equal to FGH, an a part ABE equal to FGL; therefore the remainder EBC is equal to LGH, and EB to BC as LG is to GH; but the angle BCD is equal to GHK; and a part BCE to a part GHL ; therefore the remainder ECD is equal to LHK, and the fides about the, equal angles proportional. Now, because the triangle ABE is equiangular to the triangle FGL, and the fides about the equal angles proportional, the two triangles are fimilar, and are to one another in the duplicate ratio |