Book II. Τ Η Ε Ε Ι Ε Μ Ε Ν Τ S OF E U C L I D. BOOK II. DEFINITIONS, EY I. by any two of the straight lines which contain one of the In every parallelogram, any of the parallelograms about a diame ter, together with the two complements, is called a A E Gnomon. Thus the på. rallelogram HG together with the complements AF, • FC is the gnomon, which F F K G EHC which are at the opposite angles of the parallelograms which make the gnomon.' I into any F there be two straight lines, one of which is divided 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. a. II. 1. Let A and BC be two straight lines; and let BC be divided 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;B Di E.C and to that contained by A, DE; and also to that contained by A, EC. From the point B draw : BFG at right angles to BC, and make K L H b. 3. 1. BG equal o to A; and thro' G GF A by GB, BD, of which GB is equal to A; and DL is contained d. 34. I. by A, DE, because DK, that is a 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. C. 31. I. F a straight line be divided into any two parts, the rectangies contaiņed 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 A с в Upon AB describe & the square ADEB, and AFD F E * 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. I. is the rectangle contained by BA, AC; for it is contained by Book II. 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. F 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. Upon BC describe : the square A O a. 46. 1. B CDEB, and produce ED to F, and thro' A draw 6 AF parallel to CD b. 31. 8. 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 F D E 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. 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 Ci the square of AB is equal to the squares of AC, CB and to twice the rectangle contained by AC, CB. Book II. b. 31. I. C. 29. I. I. Upon AB describe the square ADEB, and join BD, and thro' C draw CGF parallel to AD or BE, and thro' G draw HK a. 46. 1. 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 d. 5. I. and opposite angle ADB; but ADB is equal d 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, A CB e. 6. 1. and therefore the fide BC is equal e to f. 34. the fide CG. but CB is equal also f to GK, and CG to BK; wherefore the Gh K D F: E gle, wherefore GCB is a right angle; and therefore also the an. gles. 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 HE, CK-are the squares of AC, CB. and g. 43. I. because the complement AG is equal & 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 AC, CB. Wherefore if a straight line, &c. Q. E. D. Cor. From the demonstration it is manifeft, that the parallelograms about the diameter of a square are likewise squares. Book II. PROP. V. THEOR. F 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. Upon CB describe a the square CEFB, join BE, and thro' D a. 46. I. draw o DHG parallel to CE or BF; and thro' H draw KLM b. 31. I. parallel to CB or EF; and also thro' A draw AK parallel to CL or BM. and because the complement CH is equal to the comple C. 43. I. ment HF, to each of these add DM, therefore the whole A D B CM is equal to the whole L HI DF; but CM is equal d to K! M d. 36. I. AL, because AC is equal to CB; therefore also AL is equal to DF. to each of these add CH, and the whole AH E G F is equal to DF and CH. but AH is the rectangle contained by AD, DB, for DH is equal e to e. Cor. 4. . DB; and DF together with CH is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD, DB. to each of these add LG, which is equal e to the square of CD, therefore the gnomon @MG together with LG is equal to the rectangle AD, DB together with the square of CD. but the gnomon CMG and LG make up the whole figure CEFB, which is the square of CB. therefore the rectangle AD, DB together with the square of CD is equal to the square of CÃ. Wherefore if a straight line, &c. Q. E. D. |