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 .flow { margin: 0; font-size: 1em; } .flow .pagebreak { page-break-before: always; } .flow p { text-align: left; text-indent: 0; margin-top: 0; margin-bottom: 0.5em; } .flow .gstxt_sup { font-size: 75%; position: relative; bottom: 0.5em; } .flow .gstxt_sub { font-size: 75%; position: relative; top: 0.3em; } .flow .gstxt_hlt { background-color: yellow; } .flow div.gtxt_inset_box { padding: 0.5em 0.5em 0.5em 0.5em; margin: 1em 1em 1em 1em; border: 1px black solid; } .flow div.gtxt_footnote { padding: 0 0.5em 0 0.5em; border: 1px black dotted; } .flow .gstxt_underline { text-decoration: underline; } .flow .gtxt_heading { text-align: center; margin-bottom: 1em; font-size: 150%; font-weight: bold; font-variant: small-caps; } .flow .gtxt_h1_heading { text-align: center; font-size: 120%; font-weight: bold; } .flow .gtxt_h2_heading { font-size: 110%; font-weight: bold; } .flow .gtxt_h3_heading { font-weight: bold; } .flow .gtxt_lineated { margin-left: 2em; margin-top: 1em; margin-bottom: 1em; white-space: pre-wrap; } .flow .gtxt_lineated_code { margin-left: 2em; margin-top: 1em; margin-bottom: 1em; white-space: pre-wrap; font-family: monospace; } .flow .gtxt_quote { margin-left: 2em; margin-right: 2em; margin-top: 1em; margin-bottom: 1em; } .flow .gtxt_list_entry { margin-left: 2ex; text-indent: -2ex; } .flow .gimg_graphic { margin-top: 1em; margin-bottom: 1em; } .flow .gimg_table { margin-top: 1em; margin-bottom: 1em; } .flow { font-family: serif; } .flow span,p { font-family: inherit; } .flow-top-div {font-size:83%;} Exp. 1 | Hyp. 2 Concl. + Let A and BC be A B D E C the two st. lines, BC being divided in .s D and E; then A.BC o A. BD O A . DE + G! DA.EC. к L H F At. B draw BF at rt. angles to BC; make BG through D, E, and C draw DK, EL, and CH || BG, and through G, GH || BC; then o BH Os BK + DL + EH. Cons. 1 by 11. I. 2 3. I. 331, I. A: 4 Concl. Dem. 1 by Def.1&C. :: 0 BH is contained by the lines GB, BC, of which GB A; 2 Concl. ..o BH = o A.BC: 3 Def. 1&C. Also : BK is contained by GB, BD, of which GB = A; 4 Concl. .: BK O A.BD: 5 C. & 34. I. And : ADL is contained by DK, DE, of which DK = GB = A; 6 Concl. :: o DL = A.DE: 7 Sim. In like manner o EH = O A.EC: 8 Ax. 8. ..O A.BC = OS A. BD + A.DE + A.EC. 9 Recap. Wherefore, if there be two st. lines, one of which, &c. Q.E.D. COR. 2 A. 1 BC; or 3 A. } BC; or 4 A. I BC, &c. = A.BC. Sch. - The propositions of this Book may be verified by Algebra and by Arithmetic; and in doing this we shall first state the Hypothesis algebraically and numerically, and then separately give, what are denominated, the Algebraic and Arithmetical Proofs. Alg. & Arith. Hyp.-—Let A = a = 6; BC=b= 10; BD+ DE + EC=m +n + p = 5 + 3 + 2 10. = m + n + P Arith. 10 = 5 + 3 + 2 (→ a) ab = am + an + ap (~ 6) 6 x 10 = (6 x 5) + (6 x 3) + (6 x 2) or, 60 = 30 + 18 + 12 USE AND APP.-One of the methods of Demonstrating the Rule for the Multiplication of numbers depends on this proposition. In the last figure, let A represent 8, and B C 54. We cut or separate the number 54 into as many parts as there are digits : for example, 50.+ 4; each Alg. 6 part is multiplied by 8: the one part 4 x 8 = 32, and the other part 50 x 8 400. Now, all the partial products make up the whole product; therefore (4.x 8) + (50 x 8) = 54 x 8; or 32 + 400 = 432. PROP. 2. — THEOR. If a st. line be divided into any two parts, the rectangles contained by the whole st. line and each of the parts, are together equal to the square of the whole st. line. Cons.—46. I. On a given st. line to describe a square. 31. Through a given point to draw a parallel to a given st. line. DEM.—Def. 30. I. Of four-sided figures, a square is that which has all its sides equal, and all its angles rt. angles. Ax. 1. Magnitudes which are equal, &c. EXP. 1| Нур. Let st. line AB be di- D! E. vided into any two parts in .C; 2 Concl. then the OS AB. AC + AB.CB= the sq. on AB. AB; Dem, 1 by Cons. .:0 AF OS AE + CF, and A is the square on A B : 2 Def. 30. 1. also :: AE is contained by AD, AC, of which AD 3 Ax. 1. ..O AE AB. AC: 4 Cons. And CF is contained by BF, CB, of which BF -: AB; 5 Ax. 1. ..o CF AB.CB: 6 D. 3 & 5. Therefore OS AB. AC + AB. CB the square on AB. 7 Recap. If, therefore, a st. line be divided into any two Q.E.D. Alg. & Arith. Hyp.-Let AB = a units 9, and AC + CB = m + n= 5 + 4. parts, &c. Alg. Then m + n = a (x,a) :: am + an = ax a, or ao | (x 9) :. 45 + 36 9 x 9 = 81 Sch.-1. There is no necessity for the absolute construction of the rectangles, to establish the relations they express.”—LARDNER'S Euclid, p. 66. Thus, Given the line A =*k + y, to prove that A? Ak + Ay. Take the line B A ; then B. A = Bk + By = Ak + Ay. 2. The object of most of the propositions of this book is, to determine the relations between the rectangles under the parts of divided lines. We shall first confine our attention to a finite line divided into two parts." In this case there are three lines to be considered,—1st, the whole line, expressed by W; 2nd, the greater part, by P; 3rd, the less part, by p: then W? = (W.P + W. p). But the two parts may be considered as two independent lines, L, and l ; then the whole line is their sum, S; and S2 = S. L + S.l; and D being the difference, or L-1,La = L.l+L.D. USE AND APP.—Numerical Multiplication may also be proved by this process; for if a number be divided into its parts, the square of the number, which is the product of the number multiplied into itself, equals the sum of the products of each part into the undivided number. In the same way in Algebraical Equations, in which a quantity may be represented by a, and its parts by m +n; if both sides of the equation are multiplied by the quantity a, then a x a ora? = (m x a) + (n x a) = ma t na. The wood engraver substituted k for x in the figures ; hence the use of k. PROP. 3.—THEOR. If a st. line be divided into any two parts, the rectangle contained by the whole line and one of the parts, is equal to the rectangle contained by the two parts together with the square of the aforesaid part. Cong.--46. I., Pst. 2, and 31. I. DEM.—Def. 1. II., Def. 30. I., and Ax. 1. Exp. 1 Hyp. Let the st. line AB be A C B cut in. C; 2 Concl. Then O AB.BC = AC.CB + the square on BC. Cons. 1 by46.I. Pst.2 On st. line BC draw a F square CDEB, and produce ED to F; A231. I. through A draw AF | CD or BE; B B = 5 | Нур. parts, &c. Dem. 1 by Def.1.11.).:: AE is contained & 30. I. S by AB. BE, of which BE = BC; F D E 2 Ax. 1. ..O AE = AB. BC: 3 Def. 1. II. Also, ; AD is con- A ķ & 30. I. tained by AC.CI, of which CD BC; 4 Ax. 1. ..O AD AC.BC: And fig. DB is the square on CB: 6 D. 4 and 5... O AB. BC = 0 AC.BC + BC?. 7 Recap Thus, if a st. line be divided into any two Q.E.D. Alg. & Arith. Hyp.—Let A B=a=9; BC=m=6; and AC=n=3. Alg. Then a=m+n ( x m) Arith. Then 9=6+3 (~ 6) 1 ..ma=ma + mn :. 54=36 +18 Or, Let A be a line divided into k and y, and B another line=k; Then (1. II.) A . B=B.k+B.y. But (Hyp.) B=k. Therefore, B.k=k’; and B.y=k.y. Thus A. B=ka + k . y. Cor. 1. AR – B2 = (A + B). (A - B); or (k + y)2 – K2 = (k + y +k). (k + y - k) = (2k + y). y; or 81 - 36 = 45 = 15 x 3. 2. Ao – B’ is greater than (A – B)? by twice B.(A – B); or (k + y)” — ko is greater than (k + y - k) or y by 2k. (k + y - k) or 2k.y; or 81 – 36 or 45 is greater than 9 by twice 6 x 3, or 36. ☺ USE AND APP.—Multiplication of numbers may also be proved by this 3rd Prop.; for if a number, as 56, has to be multiplied by another, as 7, if the number, as 56, be separated into two parts, of which the multiplier 7 shall be one part, then on taking the square of the multiplier 7 one part, and multiplying the other part 49 by the same muitiplier, the product will equal 7 times 56. Thus 56 x 7, or 392=(7 x 7) + (49 x 7) = 49+343. PROP. 4.—THEOR. If a st. line be divided into any two parts, the square of the whole line equals the squares of the two parts, together with twice the rectangle contained by the parts. Cons.--46. I., Pst. 1, and 31. I. DEM.—29. I. If a st. line fail on two parallel st. lines, it makes the alternate angles equil, and the ext. angle equal to the int. opposite angle, and the twu interior angles equal to two rt. angles. Def. 30. I. A square has its sides equal, and its angles rt. angles. 5. I. The apgles at the base of an isosceles triangle are equal, and if the equal sides be produced, the angles on the other side of the base shall be equall. Ax. 1. Magnitudes, &c. 6. I. If two angles of a triangle be equal to one another, the sides also which subtend the equal angles shall be equal to one another. Ax. 3. If equals be taken, &c. 34. I. The opposite sides and angles of parallelograms are equal to one another, and the diagonal bisects them. 43. I. The complements of the parallelogram which are about the diameter of any parallelogram, are equal to one another. EXP. 1(Нур. Let the st. line AB be a с B divided at.C; 2 Concl, then the square on AB the squares 1 G K K on AC and CB, together with twice the O AC. CB. Cons. 1 by 46.I.Pst. 1 On AB construct the square ADEB, and D join DB; 2 31. 1. through C draw CGF A || AD or BE, and through G, HK || A B or DE; 3 Concl. then 0 AE O HF + O CK + 2 O AG. DEM. 1 byC.2 & 29.1. st. line BD falls on the || CF, AD, the ext. 1 BGC the int. L ADB : 2 C.1;D. 30.1. But fig. ADEB being a square, side AB side AD: 35. I. Ax. 1. And : LADB L ABD, and BGC LCBG: 46. I. .. side BC side CG: 5 34. I. D. 4. But GK BC, CG BK, and :: BC GC GK 6 Ax. 1. and : the fig. CGK B is equilateral. 7 C. 2, 29. I. Again, : st. line CB meets the ||s CG, BK, the 4s KBC, GCB two rt. angles : 8 D. 30, Ax.3. but K BC is a rt. angle, .. GCB is a rt. angle ;** 934. I. and .. the L8 opposite, CGB, GKB, are rt. angles ; BK; « ĐńďçăďýěĺíçÓőíÝ÷ĺéá »
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