« ΠροηγούμενηΣυνέχεια »
Exp. 1 | Hyp.
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
CH || BG, and through G, GH || BC; then o BH Os BK + DL + EH.
Cons. 1 by 11. I.
2 3. I.
Dem. 1 by Def.1&C. :: 0 BH is contained by the lines GB, BC,
of which GB A;
of which GB = A;
of which DK = GB = A;
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.
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
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.
parts in .C;
+ AB.CB= the sq.
Dem, 1 by Cons.
OS AE + CF,
and A is the square on A B :
of which AD
of which BF -: AB;
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.
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.
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.
B cut in. C; 2 Concl. Then O AB.BC =
AC.CB + the square
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;
5 | Нур.
Dem. 1 by Def.1.11.).:: AE is contained & 30. I. S by AB. BE, of which BE = BC;
of which CD BC; 4 Ax. 1. ..O AD AC.BC:
And fig. DB is the square on CB:
Arith. Then 9=6+3 (~ 6)
1 ..ma=ma + mn
:. 54=36 +18
Then (1. II.) A . B=B.k+B.y. But (Hyp.) B=k.
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.
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
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
|| 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 :
.. side BC side CG: 5 34. I. D. 4. But GK
the 4s KBC, GCB two rt. angles :