C43. I. X 43. I. Book II. equal to KN: For the same reason, PR is equal to RO; and because CB is equal to BD, and GK to KN, the rectangle B 36. 1. CK is equal o to BN, and GR to RN: But CK is equal o to RN, because they are the complements of the parallelogram is equal to BD, and that BD is CBD GK N GP, and PR to RO, the rectangle P Р RO AG is equal to MP, and PL to RF: But MP is equal to PL, because they are the complements of the parallelogram ML; wherefore AG also is equal to RF: Therefore the four rectangles E 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 recte angle 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. d Cor. 4. 2. To each of thefe add XH, which is equal d to the square of AC: Therefore four times the rectangle AB, BC, together with the square of AC, is equal to the gnomon AOH and the square XH: But the gnomon AOH and XH make up the figure AEFD which is the square of AD: Therefore four times the rectangle AB, BC, together with the square of AC, is equal to the square of AD, that is, of AB and BC added together in one straight line. Wherefore, if a straight line, &c. Q. E. D. H LF PROP. Book 11. PRO P. IX. THEOR. IF a straight line be divided into two equal, and also in to two unequal parts; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section. Let the straight line AB be divided at the point Cinto two equal, and at D into two unequal parts : The squares of AD, DB are together double of the squares of AC, CD. From the point C draw a CE at right angles to AB, and a 11, 1. make it equal to AC or CB, and join EA, EB; thro' D draw • DF parallel to CE, and through F draw FG parallel to AB; b 31. 5. and join AF: Then, because AC is equal to CE, the angle EAC is equal to the angle AEC; and because the angle C s.1. ACE is a right angle, the two others AEC, EAC together make one right angled; and they are equal to one another; d 32. 1. each of them therefore is half of a right angle. For the same E reason each of the angles CEB, EBC is half a right angle; and therefore the whole AEB is a G F right angle: And because the an. gle GEF is half a right angle, and ÈGF a right angle, for it is equalo to the interior and oppo A C D B € 29. 1. fite angle ECB, the remaining ang!e EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the fide EG equal f to the fide GF: Again, because the angle at B f 6. I. 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 side DF tof the fide DB: And be. cause AC is equal to CE, the square of AC is equal to the square of CE; therefore the squares of AC, CE are double of the square of AC: But the square of EA is equal 8 to the S 47.1, squares of AC, CE, because ACE is a right angle; therefore the square of EA is double of the square of AC: Again, becaufe EG is equal to GF, the square of EG is equal to the square of GF; therefore the squares of EG, GF are double of thc Book II. the square of GF; but the square of EF is equal to the squares of EG, GF; therefore the square of EF is double of the square h 34. I. GF: And GF is equal' to CD; therefore the square of EF is double of the square of CD: But the square of AE is likewise double of the square of AC; therefore the squares of AE, EF are double of the squares of AC, CD: And the square of AF is i 49.1. equal i to the squares of AE, EF, because AEF is a right angle; therefore the square of AF is double of the squares of AC, CD: But the squares of AD, DF are equal to the square of AF, because the angle ADF is a right angle; therefore the squares of AD, DF are double of the squares of AC, CD: And DF is equal to DB ; therefore the squares of AD, DB are double of the squares of AC, CD. If therefore a straight line, &c. Q. E. D. PRO P. X. THE O R. IF a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced. 3 II. I. Let the straight line AB be bisected in C, and produced to the point D; the squares of AD, DB ore double of the squares of AC, CD. From the point C draw a CE at right angles to AB: And make it equal to AC or CB, and join AE, EB; through E draw b 31. 1. bEF parallel to AB, and through D draw DF parallel to CE: And because the straight line Er' meets the parallels EC, FD, the € 29. I. angles CEF, EFD are equal to two right angles; and therefore the angles BEF, EFD are less than two right angles: But itraight lines which with another straight line make the interior angles d 12. Ax. upon the same lide less than two right angles, do meet d if pro. duced far enough : Therciore EB, ID ihall meet, if produced, toward B, D: Let them meet in G, and join AG:Then, becaule AC is equal to CE, the angle CLA is equal to the angle EAC: and the angle ACE is a right angle; therefore each of the f $2. I. angles CEA, EAC is half a right angles: l'or the same reason, cach e 5.1. each of the angles CEB, EBC is half a right angle; therefore Book IT. AEB is a right angle: And because EBC is half a right angle, L DBG is also f half a right angle, for they are vertically oppo- f 15.1. Ite; but BDG is a right angle, because it is equal to the al- c 29. I. tcrnate angle DCE; therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore also the side BD is equal to the side DG: Again, 8 6. I. becaute EGF is half a right angle, and that E F the angle at F is a right angle, because it is e. qual hi to the opposite h 34. I. angle ECD, the remain A ing angle FEG is half a BD right angle, and equal to the angle EGF; G wherefore also the side GF is equals to the side FE. And be. caufe EC is equal to CA, the square of EC is equal to the square of CA; therefore the squares of EC, CA are double of the square of CA: But the square of EA is equali to the squares i 47. 1. of EC, CA ; therefore the square of EA is double of the square of AC : Again, because GF is equal to FE; the square of GF is equal to the square of FE ; and therefore the squares of GF, FE are double of the square of EF: But the square of EG is equal i to the squares of GF, FE; therefore the square of EG is double of the square of EF : And EF is equal to CD; wherefore the square of EG is double of the square of CD: But it was demonstrated, that the square of EA is double of the square of AC; therefore the squares of AE, EG are double of the squares of AC, CD: And the square of AG is equal to the squares of AE, EG; therefore the square of AG is double of the squares of AC, CD: But the squares of AD, DG are equali to the square of AG; therefore the squares of AD, DG are double of the squares of AC, CD: But DG is equal to DB ; therefore the squares of AD, DB are double of the squares of AC, CD: Wherefore, if a straight line, &c. Q. E. D. PROP Book 11. PRO P. XI. PRO B. To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part. a 40. I. b 1o. I. C 3. 1. 6. 2. 47. 1. Let AB be the given straight line; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part. Upon AB describe the square ABDC; bisect b AC in E, and join BE ; produce CA to F, and make EF equal to EB; and upon AF describe a the square FGHA; AB is divided in H so, that the rectangle AB, BH is equal to the square of AH. Produce GH to K: Because the straight line AC is bifected in E, and produced to the point F, the rectangle CF, FA, together with the square of AE, is equal d to the square of EF : But EF is equal to EB; therefore the rectangle CF, FA, together with the square of AE, is equal to the square of EB: And the squares of BA, AE are equal to the F G H B E gure FK is the rectangle contained by CF, FA, for AF is equal to FG; and AD is the square of AB; therefore FK is equal to AD: Take away the common part AK, and the remainder C K D FH is equal to the remainder HD: And HD is the rectangle contained by AB, BH, for AB is e. qual to BD; and FH is the square of AH: Therefore the rectangle AB, BH is equal to the square of AH: Wherefore the straight line AB is divided in H, so that the rectangle AB, BH is equal to the square of AH. Which was to be done. PROP. |