Book II. a. 46. I. b. 31. 1. C. 29. I. d. 5. I. c. 6. I. b e A Upon AB defcribe the fquare ADEB, and join BD, and thro' C draw CGF parallel to AD or BE, and thro' G draw HK parallel to AB or DE. and because 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 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 to the fide CG. but CB is equal alfo f to GK, and CG to BK; wherefore the figure 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 an- D H C B K FE gle, wherefore GCB is a right angle; and therefore alfo the angles f CGK, GKB oppofite to thefe are right angles, and CGKB is rectangular. but it is alfo equilateral, as was demonftrated; wherefore it is a fquare, and it is upon the fide CB. for the fame reafon HF alfo is a fquare, and it is upon the fide HG which is equal to AC. therefore HF, CK are the fquares of AC, CB. and 8. 43. 1. 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 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 fquares of AC, CB and to twice the rectangle AC, CB. but HF, CK, AG, GE make up the whole figure ADEB which is the fquare of AB. therefore the fquare of AB is equal to the fquares of AC, CB and twice the rectangle AC, CB. Wherefore if a ftraight line, &c. Q. E. D. COR. From the demonstration it is manifeft, that the parallelograms about the diameter of a fquare are likewife fquares. IF PROP. V. THEO R. a ftraight line be divided into two equal parts, and and alfo into two unequal parts; the rectangle contained by the unequal parts, together with the fquare of the line between the points of fection, 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 fquare of CD, is equal to the fquare of CB. c 47 Book II. Upon CB describe the fquare CEFB, join BE, and thro' Da. 46. 1. draw DHG parallel to CE or BF; and thro' H draw KLM pa-b. 31. s. rallel to CB or EF; and alfo 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 CM is equal to the whole DF; but CM is equal AL, because AC is equal to A ୯ D B L H to K Md. 36. 1. EG F CB; therefore alfo AL is e qual to DF. to each of these add CH, and the whole AH is equal to DF and CH. but AH is the rectangle contained by AD, DB, for DII is equal toe. Cor. 4.2. DB; and DF together with CH is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD, DB. to each of thefe 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 fquare of CB. Wherefore if a ftraight line, &c. Q. E. D. Book II. 2. 46. r. b. 31. 1. c. 36. 1. PROP. VI. THE OR. IF Let the ftraight line AB be bifected in C, and produced to the point D; the rectangle AD, DB together with the fquare of CB, is equal to the fquare of CD. b Upon CD defcribe the fquare CEFD, join DE, and thro' B draw BHG parallel to CE or DF, and thro' H draw KLM parallel to AD or EF, and alfo thro' A draw AK parallel to CL or DM. and because AC is equal to CB, the rectangle AL is equal to d d. 43. 1. CH; but CH is equal to A C HF; therefore alfo AL is M the whole AM is equal to c. Cor.4. 2. DM is equal to DB. therefore the gnomon CMG is equal to the rectangle AD, DB. add to each of thefe LG, which is equal to the fquare of CB; therefore the rectangle AD, DB together with the fquare of CB is equal to the gnomon CMG and the figure LG. but the gnomon CMG and LG make up the whole figure CEFD, which is the fquare of CD; therefore the rectangle AD, DB together with the fquare of CB, is equal to the fquare of CD. Wherefore if a ftraight line, &c. Q. E. D. IF PROP. VII. THEOR. F a ftraight line be divided into any two parts, the fquares of the whole line, and of one of the parts are equal to twice the rectangle contained by the whole and that part, together with the fquare of the other part. Let the ftraight line AB be divided into any two parts in the Book II. point C; the fquares of AB, BC are equal to twice the rectangle AB, BC together with the fquare of AC. a Upon AB describe the square ADEB, and construct the figure a 46. 1. as in the preceding Propofitions. and because AG is equal to GE, b. 43. l. add to each of them CK, the whole AK is therefore equal to the c A C B G K c. Cor. 4. whole CE; therefore AK, CE are IF D PROP. VIII. THE OR. F E F a ftraight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the fquare of the other part, is equal to the fquare of the ftraight line which is made up of the whole and that part. Let the ftraight line AB be divided into any two parts in the point C; four times the rectangle AB, BC, together with the fquare of AC, is equal to the fquare of the ftraight line made up of AB and BC together. Produce AB to D fo that BD be equal to CB, and upon AD defcribe the fquare AEFD; and conftruct two figures fuch as in the preceding. Because CB is equal to BD, and that CB is equal a. 34. 3. D b. 36. 1. € 43. 1. c b Book II. to GK, and BD to KN; therefore GK is equal to KN. for the fame reafon PR is equal to RO. and because CB is equal to BD, and GK to KN, the rectangle CK is equal to BN, and GR to RN. but CK is equal to RN, because they are the complements of the parallelogram CO; therefore alfo BN is equal to GR. and the four rectangles BN, CK, GR, RN, are therefore equal to one another, and so are quadruple of one of them CK. again, because CB is equal to BD, and that BD is J. Cor. 4. 2. equal to BK, that is to CG; and CB equal to GK, that is to GP; A, becaufe CG is equal to GP, and PR M c E CB D GKN PRO HL F And it was to RO, the rectangle AG is equal X to MP, and PL to RF. but MP is equal to PL, because they are the complements of the parallelogram ML; wherefore AG alfo is equal to RF. therefore the four rectangles AG, MP, PL, RF are equal to one another, and fo are quadruple of one of them AG. demonftrated that the four CK, BN, GR, RN are quadruple of CK. therefore the eight rectangles which contain the gnomon AOH, are quadruple of AK. and because AK is the rectangle contained by AB, BC, for BK is equal to BC; four times the rectangle AB, BC is quadruple of AK. but the gnomon AOH was demonftrated to be quadruple of AK; therefore four times the rectangle AB, BC is equal to the gnomon AOH. to each of thefe add XH, which Cor. 4. 1. is equal to the fquare of AC; therefore four times the rectangle AB, BC together with the fquare of AC is equal to the gnomon AOH and the fquare XH. but the gnomon AOH and XH make up the figure AEFD which is the fquare of AD. therefore four times the rectangle AB, BC together with the fquare of AC is equal to the fquare of AD, that is, of AB and BC added together in one straight line. Wherefore if a straight line, &c. Q. E. D. |