THE ELEMENTS -OF EUCLID. BOOK II. DEFINITIONS. I. EVERY right-angled parallelogram is said to be Book II. contained by any two of the straight lines which contain one of the right angles. II. E D In every parallelogram, any of the parallelograms about H K В G 'which are at the opposite angles of the parallelograms which makes the gnomon.' 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. Book II. Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E; the rectangle con tained by the straight lines A, B 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 G BF at right angles to BC, 3.1. and make BG equal to A; * 31. 1. and through G draw GH pa- F C rallel to BC; and through D, E, C, 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, 34. 1. because DK, that is BG is equal to 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. Ir a straight line be divided into any two parts, the Let the straight line AB be divided B 46.1. Upon AB describe the square 31. 1. ADEB, and through C drawb 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, AC; for it is contained by DA, AC of which AD is equal to AB; and CE is contain D F *N. B. To avoid repeating the word contained too frequently, the rect angle contained by two straight lines AB, AC is sometimes simply called the rectangle AB, AC. ed by AB, BC, for BE is equal to AB; therefore the rect- Book II. angle 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. PROP. III, THEOR. Ir 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 aforesaid part. Let the straight line AB be divided into two parts in the point C; the rectangle AB, BC is equal to the rectangle AC, CB, together with the square of BC. Upon BC describe the square CDEB, and produce 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 C a 46 1. B b 31. 1. E CD is equal to BC; 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. PROP. IV. THEOR. Ir 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. a 46. 1. Upon AB describe a the square ADEB, and join BD, and through C draw CGF parallel to AD or BE, and through 31. 1. G draw HK parallel to AB or DE: And because CF is parallel to AD, and BD falls upon them, the exterior angle € 29. 1. d 5. 1. C A BOOK II. BGC is equal to the interior and opposite angle ADB; but ADB 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 6. 1. the side BC is equal to the side 34. 1. CG: But CB is equalf also to GK, HE D G K F and CG to BK; wherefore the figure CGKB is equilateral: It is likewise rectangular; for CG is parallel to BK, and CB meets them; the angles KBC, GCB are therefore equal to two right angles; and KBC is a right angle; 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 comple43. 1. ment AG is equals to the complement GE, and that AG is the rectangle 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 AČ, 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. BOOK II. PROP. V. THEOR. If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line. Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts at the point D; the rectangle AD, DB, together with the square of CD, is equal to the square of CB. b Upon CB describe the square CEFB, join BE, and * 46. 1. through D drawb DHG parallel to CE or BF; and through 31. 1. H draw KLM parallel to CB or EF; and also through A draw AK parallel to CL or BM: And because the complement CH is equal to the complement HF, to each of 43. 1. these add DM; therefore the whole CM is equal to the whole DF; butCM is equald d 36. 1. C D B whole AH is equal to DF E G F and CH: but AH is the ་ ་ From this proposition it is manifest, that the difference of the squares of two unequal lines AC, CD, is equal to the rectangle contained by their sum and difference. |