EXP. 1| Hyp. 2 Concl. T CONS.1 by 11. I. 2 3. I. At. B draw BF at rt. angles to BC; make BG = A: through D, E, and C draw DK, EL, and CH |s BG, and through G, GH || BC; 3 31. I. 4 Concl. DEM. 1 by Def.1&C. BH is contained by the lines GB, BC, 3 Def. 1 & C. Also BK is contained by GB, BD, 5 C. & 34. I. And DL is contained by DK, DE, 6 Concl. 7 Sim. 8 Ax. 8. 9 Recap. In like manner EHA.EC: .. A.BC =□s A. BD + A. DE + A. EC. Wherefore, if there be two st. lines, one of which, &c. COR. 2 A.BC; or 3 A. = Q.E.D. BC; or 4 A. 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. 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 BC 54. We cut or separate the number 54 into as many parts as there are digits: for example, 50 + 4; each part is multiplied by 8: the one part 4 × 8: = 32, and the other part 50 x 8 400. Now, all the partial products make up the whole product; therefore (4 × 8) + (50 × 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. DEM. 1 by Cons. then s DC + EB the square DB. AF = SAE + CF, and A is the square on AB: 2 Def. 30. I. also AE is contained by AD, AC, 3 Ax. 1. 4 Cons. 5 Ax. 1. 6 D. 3 & 5. 7 Recap. square on AB. If, therefore, a st. line be divided into any two parts, &c. Alg. & Arith. Hyp.-Let AB: = a units = Q.E.D. 9, and AC + CB = m + n= 5 +4. Alg. Then m + n = a (x,a) SCH.-1. = Arith. Then 5 + 4 = 9 am + an = a × a, or aa | (× 9) .. 45 + 36 9 × 981 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 A2 = 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 W2 = (W . P + W. p). = But the two parts may be considered as two independent lines, L, and 7; then the whole line is their sum, S; and S2 S.LS. and D being the difference, or L-1, L2 = L. 1+ 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 or a2 = (m x a) + (n × a) = ma + 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. CONS.1 by46.I.Pst.2 On st. line BC draw a F 2 31. I. square CDE B, and produce ED to F; through A draw AF || CD or BE; A B L 5 Hyp. = And fig. DB is the square on CB: 6 D. 4 and 5. .. □ AB. BC □ AC. BC + BC2. 7 Recap. = Thus, if a st. line be divided into any two parts, &c. Alg. & Arith. Hyp.-Let A B=a=9; BC=m=6; and AC=n=3. Alg. Then a m+n (xm) ..ma=m2 + mn Arith. Then 9=6+3 (x6) .. 54=36+18 Or, Let A be a line divided into k and y, and B another line=k; Q.E.D. Therefore, B. k=k'; and B. y=k. y. Thus A.B=k2+k. y. COR. 1. A2 - B2 = (A + B). (A−B); or (k + ?1)2 − k2 = (k + y + k). (k+y-k) = (2k + y). y; or 81-36-4515 x 3. 2. A-B is greater than (AB) by twice B (AB); or (k + y)-k2 is greater than (k + y −k)2 or y3 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 × 7) + (49 × 7)=49+343. 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 equal, and the ext. angle equal to the int. opposite angle, and the two interior angles equal to two rt. angles. Def. 30. I. A square has its sides equal, and its angles rt. angles. 5. I. The angles 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 equal. 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. the fig. CGKB is equilateral. 7 C. 2, 29. I. Again, st. line CB meets the ||s CG, BK, 8 D. 30, Ax.3. = two rt. angles : the 48 KBC, GCB but K BC is a rt. angle, .. GCB is a rt. angle; and the s opposite, CGB, GKB, are rt. angles; |