If the square described upon BC, one of the sides of the Book I. triangle ABC, be equal to the squares upon the other sides BA, AC, the angle BAC is a right angle. a D From the point A draw AD at right angles to AC, and a 11. 1. make AD equal to BA, and join DC: Then because DA is equal to AB, the square of DA is equal to the square of AB. To each of these add the square of AC; therefore the squares of DA, AC, are equal to the squares of BA, AC: But the square of DC is equal to the squares of DA, AC, because DAC is a right angle; and the square of BC, by hypothesis, is equal to the squares of BA, AC; therefore b the square of DC is equal to the square B A C b 47. 1. of BC; and therefore also the side DC is equal to the side BC. And because the side DA is equal to AB, and AC common to the two triangles, DAC, BAC, the two DA, AC are equal to the two BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle c 8. 1. BAC: But DAC is a right angle; therefore also BAC is a right angle. Therefore," if the square," &c. C Q. E. D. THE ELEMENTS OF EUCLID. BOOK II. DEFINITIONS. I. Book II. EVERY right angled parallelogram, (or rectangle,) is said to be contained by any two of the straight lines which contain one of the right angles. II. In every parallelogram, any of the parallelograms about a dia meter, together with the two complements, is call- Thus the 'parallelogram HG, to 'angles of the parallelograms which make the gnomon.' PROP. I. THEOR. If there be two straight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. B DEC Let A and BC be two straight lines; and let BC be divid- Book II. ed into any parts in the points D, E; the rectangle contained by the straight lines A, BC is equal to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC. From the point B draw BF G at right angles to BC, and make BG equal to A; and through G draw GH parallel F b to BC, and through D, E, C, d a 11. 1. KL H b 3. 1. A c 31. 1. draw DK, EL, CH parallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH; and BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A; and BK is contained by A, BD, for it is contained by GB, BD, of which GB is equal to A and DL is contained by A, DE, because DK, that is BG, is equal to a 34. 1. A; and in like manner the rectangle EH is contained by A, EC: Therefore the rectangle contained by A, BC is equal to the several rectangles contained by A, BD, and by A, DE; and also by A, EC. Wherefore, "if there be two straight lines," &c. Q. E. D. PROP. II. THEOR. If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square of the whole line. Let the straight line AB be divided into any two parts in the point C; the A rectangle contained by AB, BC, together with the rectangle * AB, AC, shall be equal to the square of AB. Upon A B describe the square ADEB, and through C draw CF, parallel to AD or BE; then AE is equal to the rectangles AF, CE; and AE is the square of AB; and AF is the rectangle contained by BA, D св FE N. B.-To avoid repeating the word contained too frequently, the rectangle contained by two straight lines AB, AC is sometimes simply called the rectangle AB, AC. a 46. 1. b 31. 1. Book II. AC; for it is contained by DA, AC, of which AD is equal to AB; and CE is contained by AB, BC, for BE is equal to AB; therefore the rectangle contained by AB, AC, together with the rectangle AB, BC, is equal to the square of AB. If, therefore, a straight line," &c. Q. E. D. a 46. 1. b 31. 1. 66 PROP. III. THEOR. If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the foresaid part. Let the straight line AB be divided into any two parts in the point C; the rectangle AB, BC is equal to the rectangle AC, CB, together with the square of BC. АС B Upon BC describe the square CDEB, and produce A ED to F, and through A draw b AF parallel to CD or BE; then the rectangle AE is equal to the rectangles AD, CE; and AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is equal to BC; and AD is contained by AC, CB, for CD is equal to CB; and DB is the square of BC; therefore the rectangle AB, BC is equal to the rectangle AC, CB together with the square of BC. "If, therefore, a straight line," &c. Q. E. D. FD PROP. IV. THEOR. E If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts. Let the straight line AB be divided into any two parts in C; the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB. c a b d 5. 1. Upon AB describe the square ADEB, and join BD, and Book II. through C draw CGF parallel to AD or BE, and through G draw HK parallel to AB or DE: and because CF is paral- a 46. 1. lel to AD, and BD falls upon them, the exterior angle BGC 31. 1. is equal to the interior and opposite angle ADB; but ADB c 29. 1. is equal to the angle ABD, because BA is equal to AD, being sides of a square; wherefore the angle CGB is equal to the angle GBC; and therefore the side BC is equal to the side CG: But CB is equal also to GK, and CG to BK; H wherefore the figure CGKB is equilateral: It is likewise rectangular; e A C B G K e 6. 1. f 34. 1. for CG is parallel to BK, and CB F E wherefore GCB is a right angle; and therefore also the angles CGK, GKB opposite to these, are right angles, and CGKB is rectangular: But it is also equilateral, as was demonstrated; wherefore it is a square, and it is upon the side CB; for the same reason HF also is a square, and it is upon the side HG, which is equal to AC: Therefore HF, CK are the squares of AC, CB; and because the complement AG is equal to the complement GE, and that AG is the rectangle g 43. 1. contained by AC, CB, for GC is equal to CB; therefore GE is also equal to the rectangle AC, CB: wherefore AG, GE are equal to twice the rectangle AC, CB: and HF, CK are the squares of AC, CB; wherefore the four figures HF, CK, AG, GE are equal to the squares of AC, CB, and to twice the rectangle AC, CB: But HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB: Therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle AC, CB. Wherefore, "if a straight line,” &c. Q. E. D. COR. From the demonstration, it is manifest, that the parallelograms about the diameter of a square are likewise squares. |