THE ELEMENT S OF EUCLI D. воок II. DEFINITIONS. I. EVERY right angled parallelogram is faid to be contained by any two of the ftraight lines which contain one of the right angles. II. In every parallelogram, any of the parallelograms about a diameter, together with the two complements,is called A 6 Thus the a Gnomon. 'parallelogram HG,toge there with the comple'ments AF, FC, is the gnomon, which is more ' briefly expreffed by the E D ⚫ letters AGK, or EHC, B 'which are at the oppofite 'angles of the parallelograms which make the gnomon.' IF PROP. I. THEOR. F there be two straight lines, one of which is divided into any number of parts; the rectangle contained by the two ftraight lines, is equal to the rectangles contained by the undivided line, and the feveral parts of the divided line. Book II. Book II. II. F. b 3. I. C 31. I. 814. I. 346, T. bai. I. DEC Let A and BC be two ftraight lines; and let BC be divided From the point B draw a BF and through D, E, C, draw c DK, F EL, CH parallel to BG; then KL H A 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, becaufe DK, that is d 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 feveral rectangles contained by A, BD, and by A, DE; and alfo by A, EC. Wherefore, if there be two ftraight lines, &c. Q. E. D. IF PROP. II. THEOR. Fa ftraight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the fquare of the whole line. Let the ftraight line AB be divided into * Upon AB describe a the fquare ADEB, C B FE N. B. To avoid repeating the word contained too frequently, the re&angle contained by two straight lines AB, AC is sometimes fimply called the rectangle AB,AC. 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 fquare of AB. If there fore a straight line, &c. Q. E. D. IF PROP. III. THEOR. F a ftraight 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 fquare of the forefaid 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 fquare of BC. Upon BC defcribe a the fquare A C 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 CD. is equal to BC; and DB is the square of BC; therefore the rectangle AB, D Book II. یم Ba 46. I. b 31. I. BC is equal to the rectangle AC, CB together with the square of BC. If therefore a straight, &c. Q. E. D. IF PROP. IV. THEOR. F a ftraight line be divided into any two parts, the fquare of the whole line is equal to the fquares of the two parts, together with twice the rectangle contained by the parts. Let the ftraight line AB be divided into any two parts in C; the fquare of AB is equal to the fquares of AC, CB, and to twice the rectangle contained by AC, CB, D 3 Upon A D C B G K F E Upon AB describe a the fquare ADEB, and join BD, and through C draw b CGF parallel to AD or BE, and through G draw HK parallel to AB or DE: And becaufe CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal to the interior and oppofite angle ADB ; but ADB is equal d to the angle ABD, because BA is equal to AD, being fides of a fquare; wherefore the angle CGB is equal to the angle GBC; and therefore the fide BC is equal e to the fide CG: But CB is equal f alfo to GK, and CG to BK; wherefore the fi- H gure CGKB is equilateral: It is likewife 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 alfo the angles f CGK, GKB oppofite to these, are right angles, and CGKB is rectangular: But it is alfo equilateral, as demonstrated; wherefore it is a square, and it is upon the fide CB: For the fame reafon HF also is a square, and it is upon the fide HG, which is equal to AC: Therefore HF, CK are the squares of AC, CB; and because the complement AG is equal g to the complement GE, and that AG is the rectangle contained by AC, CB, for GC is equal to CB; therefore GE is alfo equal to the rectangle AC, CB; wherefore AG, GE are equal to twice the rectangle AC, CB: And HF, CK are the fquares 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 fquares of AC, CB and twice the rectangle AC, CB. Wherefore if a straight line, &c. Q. E. D. COR. From the demonftration, it is manifest that the pa rallelograms about the diameter of a square are likewise squares. PROP 1 Book II. I' PROP. V: THEO R. F a ftraight line be divided into two equal parts, and alfo into two unequal parts; the rectangle contained by the unequal parts, together with the fquare of the line between the points of section, is equal to the fquare of half the line, Let the ftraight 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 fquare of CB. Upon CB defcribe a the fquare CEFB, join BE, and through a 46. I. D draw b DHG parallel to CE or BF; and through H draw b 31. 1. KLM parallel to CB or EF; and alfo through A draw AK parallel to CL or BM: And because the complement CH is equal to the complement HF, to each of these add DM; c 43. I. therefore the whole CM is equal to the whole DF; but CM is equal d to AL, because AC is equal to K A C D B L H d 36. I M CB; therefore alfo AL is equal to DF. To each of thefe add CH, and the whole AH is equal to DF and CH: But AH is the AD, DB, for DH, is equal rectangle contained by e e to DB; and DF together with CH is the gnomon CMG; Cor. 4.2. therefore the gnomon CMG is equal to the rectangle AD, DB: To each of these add LG, which is equal to the fquare of CD; therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the fquare of CD: But the gnomon CMG and LG make up the whole figure CEFB, which is the fquare of CB: Therefore the rectangle AD, DB, together with the fquare of CD, is equal to the square of CB. Wherefore, if a ftraight line, &c. Q. E. D. From this propofition it is manifeft, that the difference of the fquares of two unequal lines AC, CD, is equal to the rectangle contained by their fum and difference.` |