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, AC. Recite (a) p. 46; (d) ax. 7; Q. E. D. (b) p. 31; (e) ax. 2; (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, AČ), 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). 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. H C 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. 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 D C C K L 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. Recite (a) p. 11; (c) p 31; (b) p. 3; (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 A AC, CB. Describe on AB the square AE (a); draw CF parallel to AD (b). 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; 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 C (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. Q. E. D. B Recite (a) p. 46; (b) pos. 2; (c) p. 31-all of b. 1. 4 Th. The square of a straight line (AB) is equal to the squares of any two parts (AC, CB) into which it may be divided, and two rectangles of the parts Constr. Upon AB describe the square AE (a); A join BD; draw CF parallel to AD, meeting BD in G: draw HGK parallel to AB (b). H Argument. The angles CGB, ADB, as exterior and interior, are equal (c); and ADB equals ABD, from the isosceles (d); therefore, ABD equals CGB (e), and the sides CG, CB are equal (ƒ), and their opposites BK, GK are equal (g). CK is D therefore a square (h) on CB. св F E For similar reasons, HF is a square on HG, which is equal to AC. It is also plain that AG is the rectangle of AC and CG, or CB; also, that GE is the rectangle of GF and GK, which are equal to AC and CB. Now the two squares and two rectangles make up the square of AB. Wherefore, the square, &c. (b) p. 31; Q. E. D. (c) p. 29; (d) p. 5; (g) p. 34-all of b. 1: Recite (a) p. 46; (e) ax. 1; (h) def. 30 of b. 1, and 1 of b. 2. Cor. The rectangles about the diameter of a square are squares. Scholium. A line or other magnitude divided into two parts, is called a binomial, the properties of which are very remarkable. 5 Th. If a straight line (AB), be divided into two equal parts (in C), and two unequal parts (in D), the square of half the line is equal to the rectangle of the unequal parts and the square of the line between the sectional points. Upon CB, half the line, describe the A square CF (a); join BE; through D, draw DG parallel to BF (b), meeting BE in H; through H, draw ML parallel to AB, and produce it to meet AK drawn parallel to CE. K L B H M E C F The square CF equals the gnomon CMG and the square LG. But the gnomon is equal to the rectangle AH; for the complements CH, HF are equal (c), and with DM added to each, CM equals DF; but CM equals AL, (d); therefore DF equals AL, (e). Again, AH is the rectangle of the unequal parts AD, DB; for DH equals DB (f): LG is also the square of LH, or CD, the line between the sectional points. Therefore, the square of CB, is equal to the rectangle of AD, DB and the square of CD. Wherefore, if a straight line, &c. Q. E. D. Recite (a) p. 46; (b) p. 31; (c) p. 43; (e) ax. 1-all of b. 1; (d) p. 36; (f) cor. p. 4 of b. 2. 6 Th. If a straight line (AB) be bisected (in C), and produced to any point (D); the square of the line (CD), composed of the half and the part produced, is equal to the rectangle of the whole line AD and its produced part, (BD), with the square of (CB), half the line bisected. Upon CD describe the square CF (a); join DE; through B, draw BG parallel to DF, and meeting the diameter in H; through H, draw ML parallel to AB, and produce it K to meet AK drawn parallel to CE (b). The square CF is equal to the gnomon CMG and the square LG: this is evident. AM: for the rectangles AL and CH are equal (c); also CH and HF (d); therefore AL equals HF (e). Hence the square CF equals the rectangle AM and the square LG. But AM is the rectangle of the whole line AD and its produced part BD: for DM or BH equals BD (ƒ). Also, LG is the square of LH, or CB, half the bisected line. Wherefore, the square of CD equals the square of CB and the rectangle of AD and DB. Q. E. D. 7 Th. The two squares described upon a straight line (AB) and a part of that line (BC) are equal to two rectangles of the line and that part, together with the square of the other part (AC). Constr. Upon AB describe the square AE (a) join BD; through C draw CF parallel to BE, and cutting the diameter in G; through G draw HK H parallel to AB (b). The squares AE, CK are equal to the rectangles AK, CE and the square HF. This is evident. Now AE is the square of AB, and CK is the square of BC. Also, AK is the rectangle of AB A D F K and BK, or BC; and CE is the rectangle of BC and BE, or BA; and HF is the square of HG, or AC. Wherefore, the two squares described, &c. Q. E. D. Cor. Hence the sum of the squares of any two lines, is equal to twice the rectangle of the two with the square of their difference |