44 P. To a given straight line (AB) to apply a parallelogram equal to a given triangle (C), having an angle equal to a given rectilineal angle. Upon half the base of the given triangle C, make a parallelogram GBEF, equal to C (a); and let its angle GBE equal the given angle D (b); let AB, the given straight line, be the produced part of EB; through B produce GB to M; through A draw HL parallel to GM; produce FG to H; through B draw HK; produce FE to K; make KL parallel to EA. H M B The complements BL and BF are equal (c); but BF was made equal to the given triangle C; therefore, BL is equal to C (d); and it is applied to the given straight line AB, having one of its angles ABM, equal to the opposite vertical angle GBE, which was made equal to the given angle D. Therefore the thing required has been done. (b) p. 23; (c) p. 43; 45 P. To describe a parallelogram equal to a given rectilineal figure (ABCD), having an angle equal to a given rectilineal angle (P); and to apply the parallelogram to a given straight line (CM). Produce the side CB of the given rectilineal figure indefinitely to E (a); join DB (b), and parallel to it, through A, draw AE (c); join DE (b). Again, bisect CE in F (d); upon CF describe E the parallelogram CFGH, equal to the triangle DEC, and having an angle FCH equal to the given angle P (e). DG BF M N Also, let CM, the line to which the parallelogram is to be applied, be in the same straight line with HC; produce FC to O; through M draw LN parallel to FO (c); produce GF to meet NL; through C draw LK; produce GH to meet LK; parallel to HCM draw KÖN (c.) 1. The triangles DAB, DEB, on the same base DB, and between the same parallels DB, AE, are equal to one another (f); to each add the triangle DBC; therefore, the triangle DEC is equal to the given rectilineal figure ABCD (g). 2. But the parallelogram CFGH was made equal to the triangle DEC; therefore CFGH is equal to the given rectilineal figure ABCD (h), and its angle FCH is equal to the given angle P. 3. The complements FH and MO are equal (i); and the angle MCO is equal to its vertical and opposite angle FCH, and therefore to the given angle P (k). Wherefore to CM is applied a parallelo gram equal to a given rectilineal figure, having an angle equal to a given rectilineal angle. Recite (a) pos. 2; (d) p. 10; (e) p. 42; (f) p. 37; (g) ax. 4; (h) ax. 1; (i) p. 43; (k) p. 15; NOTE. This demonstration includes the corollary, as given in Simson's and Playfair's edition. 46 P. To describe a square upon a given straight line (AB.) Constr. From the point A draw AC at right an- c gles to AB (a); make AD equal to AB (b); draw DE parallel to AB, and BE parallel to AD (c). 1. Of the equal angles. The angle A is a right angle; and, because of the parallels, the two interior angles, A and D, are equal to two right angles (d); also, for the same reason, A equals B, and B equals E. Therefore, the figure has four right angles. B 2. Of the equal sides. The sides AB, AD are made equal; and because the opposite sides of a parallelogram are equal (e), DE is equal to AB, and BE to AD. Hence also, the figure has four equal sides; and is therefore a square (ƒ). Wherefore, the required square has been described. Recite (a) p. 11; (b) p. 3; (c) p. 31; 47 Th. In any right-angled triangle (ABC), the square upon the side (BC), subtending the right angle, is equal to the sum of the squares of the sides (AB, AC), containing the right angle. Construction. On BC describe the square BE, on AB the square BG, on AC the square AK (a); draw AL parallel to BD, or CE (b); join AD, CF; also AE, BK. Dem. Because of the adjacent right angles at A, the lines BA, AH, and CA, AG, are straight lines (c), the one parallel to BF, the other to CK. The triangle BCF is upon the side BF, and between the same parallels as the square BG, of which it is equal to the half: H Also, the triangle ABD is upon the side BA, and between the same parallels as the parallelogram BL, of which it is equal to the half (d). But the triangles BCF, ABD are equal; having two sides BF, BC, in the one, equal to two sides, BA, BD, in the other; and each of the two containing the angle ABC and a right angle (e). Therefore, the half of the square BG, is equal to the half of the parallelogram BL; hence the square is equal to the parallelogram (ƒ). In the same way, it may be shown that the square CH, is equal to the parallelogram CL. Therefore, the two parallelograms BL, CL, viz. the square BE, are equal to the two squares, BG, CH. Wherefore, the square upon BC is equal to the sum of the squares upon AB, AĆ. Recite (a) p. 46; (d) ax. 7; Q. E. D. (b) p. 31; (c) p. 14; 48 Th. If the square upon one side (BC) of a triangle be equal to the squares upon the other two sides (AB, AC), the angle contained by these two sides shall be a right angle. From the point A, draw AD at right angles to AC (a); and make AD equal to AB; join CD. Then because AD equals AB, their squares also are equal;-to each of them add the square of AC; therefore the squares of AD, AC equal the squares of AB, AC (b). C But the square of CD equals the squares of AD, AC (c), because DAC is a right angle: and the square of BC was supposed to be equal to the squares of AB, AC. Therefore, the squares of BC and CD are equal, and BC is equal to CD. And because the sides AD, AB are equal, and AC common to the two triangles, and the bases BC, CD also equal; therefore the angle BAC is equal to the angle DAC (d): but DAC is a right angle; therefore BAC is also a right angle. Wherefore, if the square, &c. (b) ax. 2; (d) p. 8-all of b. 1. Q. E. D. BOOK SECOND. Definitions. THE multiplication or division of magnitudes makes no change in their species; and magnitudes of different kinds cannot be united in additions: a part taken from a magnitude is of the same species as the whole. The units of a line are therefore lines;—of a superficies, areas;-of a solid, solids;-of an angle, angles. Any produced magnitude is therefore a multiple of its root or basis; or of the unit which measures the basis. 1. A rectangle is a superficies contained under two straight lines at right angles to each other: if the lines are equal, the figure is a square; if unequal, the figure is an oblong; and in either case, it is a right angled parallelogram. 2. A gnomon is the sum of the two complements (p. 43,) and one of the parallelo- B grams about the diameter of a rectangle. F C K NOTE 1. The operative signs +, −, X, ÷,=, are sometimes used to express the sum, difference, product, quotient and equality of magnitudes; and the parentheses () unite two or more magnitudes in one. A point is often used to denote multiplication instead of the oblique cross. NOTE 2. Because a square has two equal dimensions, the figure 2 placed over one of them saves repetition; thus, AB? is the square of AB; but in the case of a rectangle of different dimensions, both must be expressed; thus, ABX BC, or AB.BC, is the rectangle or product of AB multiplied into BC. NOTE 3. Lines used numerically, as multiplier and multiplicand, are called coefficients of each other, or co-factors. Propositions. 1 Th. The rectangle contained under two straight lines (A and BC), is equal to the several rectangles contained under one of them and all the parts of the other. Let A be a whole line and BC a divided one, as in the points D, E. Construction. Draw BG at right angles to BC (a), and make it equal to A (b); draw GH parallel to BC (c), and DK, EL, CH parallel to BG. The opposite parallels are equal (d). Argument. Now it is obvious that the B K A rectangle BH contains the three rectangles BK, DL, EH; and BH is the rectangle of BG and BC: and, since BG, DK, EL are each of them equal to A (d), therefore AXBD equals BK, AXDE equals DL, and AXEC equals EH; but BD, DE, EC are all the parts of BC: therefore, the rectangle of A and BC is equal to the rectangles of A and all the parts of BC. Wherefore, the rectangle contained, &c. (b) p. 3; Recite (a) p. 11; (c) p 31; (d) p. 34-all of book 1. Q. E. D. 2 Th. The square of a straight line is equal to two rectangles contained under that line and any two parts into which it may be divided. Let AB be a straight line divided into the parts AC, CB. Describe on AB the square AE (a); draw CF parallel to AD (6). Now the square AE equals the two rectangles AF and CE: but AE is the square of AB (c); and AF is the rectangle of AD and AC, and CE is the rectangle of CF and CB. Also, since AE is a square, AD, or CF is equal to AB, and AC, CB are the parts. Therefore, the square of a straight line, &c. Recite (a) p. 46; (b) p. 31-all of b. 1; (c) def. 30, b. 1, and 1 of b. 2. C FE Q. E. D. 3 Th. The rectangle contained by a straight line (AB) and (BC) one of two parts into which it is divided, is equal to the square of that part and the rectangle of the two parts. Constr. Upon BC describe the square BD A (a); produce ED F (b) to meet AF drawn parallel to CD, or BE (c). Then, since BE and BC, sides of a square, are equal, AE is the rectangle of AB and BC; and it contains the square of BC with the rectangle of AC and CD, or CB, and is equal to them. Wherefore, the rectangle contained, &c. (c) p. 31-all of b. 1. |