Book II. a. 46. I. c. 36. 1. d. 43. I. IF PROP. VI. THEOR. Fa ftraight line be bifected, and produced to any point; the rectangle contained by the whole line. thus produced, and the part of it produced, together with the fquare of half of the line bisected, is equal to the fquare of the ftraight line which is made up of the half and the part produced. 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. A B D L H K M Upon CD describe the square CEFD, join DE, and thro' B b. 31. 1. drawb BHG parallel to CE or DF, and thro' H draw KLM parallel to AD or EF, and also thro' A draw AK parallel to CL or DM. and because AC is equal to CB, the rectangle AL is equal to CH; but CH is equal d to HF; therefore alfo AL is equal to HF. to each of thefe add CM, therefore the whole AM is equal to the gnomon CMG. and AM is the rectangle contained by AD, DB, for e. Cor. 4. 2. DM is equal e to DB. therefore the gnomon CMG is equal to the rectangle AD, DB. add to each of these LG, which is equal to the fquare of CB; therefore the rectangle AD, DB together with the square 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 square of CD. Wherefore if a ftraight line, &c. Q. E. D. I PROP. VII. E THEOR. G F a ftraight line be divided into any two parts, the squares 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 point C; the fquares of AB, BC are equal to twice the rectangle AB, BC together with the square of AC. b 1 Book II. Upon AB describe a the square ADEB, and construct the figure a. 46. I. as in the preceding Propofitions. and because AG is equal to GE, b. 43. I. add to each of them CK, the whole AK is therefore equal to the equal to the square of AC; therefore the gnomon AKF together with the fquares CK, HF is equal to twice the rectangle AB,,BC and the fquare of AC. but the gnomon AKF together with the fquares CK, HF make up the whole figure ADEB and CK, which are the fquares of AB and BC. therefore the fquares of AB and BC are equal to twice the rectangle AB, BC together with the square of AC. Wherefore if a straight line, &c. Q. E. D. Fa 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 straight line AB be divided into any two parts in the point C; four times the rectangle AB, BC, together with the square of AC, is equal to the fquare of the straight 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 a. 34. 2, D Book II. to GK, and BD to KN; therefore GK is equal to KN. for the fame reason PR is equal to RO. and because CB is equal to BD, C. 43. I. C b. 36. 1. 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 C. I. CB equal to GK, that is to GP; A c X E CB_D GK N P zo HL F to RO, the rectangle AG is equal to MP, and PL to RF. but MP is c. 43. 1. 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 so are quadruple of one of them AG. And it was demonstrated 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 demonstrated to be quadruple of AK; therefore four times the rectangle AB, BC is equal to the gnomon AOH. to each of these add XH, which d. Cor. 4. 2. 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. d PROP. IX. THEOR. Book II. IF F a ftraight line be divided into two equal, and also into two unequal parts; the fquares of the two unequal parts, are together double of the fquare of half the line, and of the fquare of the line between the points of fection. Let the ftraight line AB be divided at the point C into two equal, and at D into two unequal parts. the fquares of AD, DB are together double of the squares of AC, CD. c ს From the point C draw a CE at right angles to AB, and make it a. 11. 1. equal to AC or CB, and join EA, EB; thro' D draw ↳ DF parallel b. 31. 1. to CE, and thro' F draw FG parallel to AB; and join AF. then because AC is equal to CE, the angle EAC is equal to the angle c. 5. I. AEC; and because the angle ACE is a right angle, the two others AEC, EAC together make one right angle ; and they are equal d. 32. I. to one another; each of them therefore is half of a right angle. e f gle EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the fide EG equal to the fide GF. again, f. 6. 1. because the angle at B is half a right angle, and FDB a right angle, for it is equal to the interior and oppofite angle ECB, the remaining angle BFD is half a right angle; therefore the angle at B is equal to the angle BFD, and the fide DF to f the fide DB. and because AC is equal to CE, the fquare of AC is equal to the fquare of CE; therefore the fquares of AC, CE are double of the square of AC. but the square of EA is equal to the fquares of AC, CE, g. 47. I. because ACE is a right angle; therefore the fquare of EA is double of the fquare of AC. again, because EG is equal to GF, the square of EG is equal to the fquare of GF; therefore the fquares of EG, GF are double of the fquare of GF; but the fquare of EF is Book II. equal to the fquares of EG, GF; therefore the fquare of EF is double of the fquare GF. and GF is equal b to CD; therefore the h. 34. I. g. 47. I. a. II. I. b. 31. I. C. 29. I. fquare of EF is double of the Ε G F A CD B AE is likewife double of the fquare PROP. X. THEOR. [Fa ftraight line be bifected, and produced to any point, the fquare of the whole line thus produced, and the fquare of the part of it produced are together double of the fquare of half the line bisected, and of the square of the line made up of the half and the part produced. Let the ftraight line AB be bifected in C, and produced to the point D; the fquares of AD, DB are double of the fquares of AC, CD. c From the point C draw a CE at right angles to AB, and make it equal to AC or CB, and join AE, EB; thro' E draw b EF parallel to AB, and thro' D draw DF parallel to CE. and because the ftraight line EF meets the parallels EC, FD, the angles CEF, EFD are equal to two right angles; and therefore the angles BEF, EFD are lefs than two right angles. but ftraight lines which with another straight line make the interior angles upon the fame d. 12. Ax. fide less than two right angles, do meet & if produced far enough. therefore EB, FD shall meet, if produced, toward BD. let them meet in G, and join AG. then because AC is equal to CE, the angle CEA is equal to the angle EAC; and the angle ACE is a right angle; therefore each of the angles CEA, EAC is half a e. 5. I. e |