COR. Hence, four times the rectangle contained by two lines Book II. AB, BC, together with the fquare of their difference AC, i equal to the fquare of their fum AD. IF PROP. IX. THEOR. Fa 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 fquares of AC, CD. From the point C draw CE at right angles to AB, and a 11. 1. make it equal to AC or CB, and join EA, EB; through D draw e d C A E C 5. 1. d 32. I. C D € 29. T. DF parallel to CE, and through F draw FG parallel to AB; b 31. 1. and join AF: Then, becaufe AC is equal to CE, the angle EAC is equal to the angle AEC; and because the angle ACE is a right angle, the two others AEC, EAC together make one right angle ; and they are equal to one another; each of them therefore is half of a right angle. For the fame reafon each of the angles CEB, EBC is half a right angle; and therefore the whole AEB is a right angle: And because the angle GEF is half a right angle, and EGF a right angle, for it is equal to the oppofite angle ECB, the remaining angle 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, becaufe f 6. 1. the angle at B is half a right angle, and FDB a right angle, for it is equal to the 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 the fide DB. And because AC is equal to CE, the square of AC is equal to the square of CE; therefore the fquares of AC, CE are double of the fquare of AC: But the fquare of EA is equal to the fquares of AC, g 47. 1. CE, because ACE is a right angle; therefore the fquare of EA is double of the square of AC: Again, because EG is equal to GF, the fquare of EG is equal to the fquare of GF; therefore the fquares of EG, GF are double of the fquare of GF: but the square of EF is equal to the fquares of EG, GF; therefore i 47. 1. the fquare of EF is double of the fquare GF: And GF is equal e f to CD; therefore the fquare of EF is double of the fquare of h 34. I. CD: i 47. 1. BOOK I. CD: But the fquare of AE is likewife double of the fquare of AC; therefore the fquares of AE, EF are double of the fquares of AC, CD: And the fquare of AF is equal to the fquares of AE, EF, because AEF is a right angle; therefore the fquare of AF is double of the fquares of AC, CD: But the fquares of AD, DF are equal to the fquare of AF, because the angle ADF is a right angle; therefore the fquares of AD, DF are double of the fquares of AC, CD: And DF is equal to DB; therefore the fquares of AD, DB are double of the fquares of AC, CD. If therefore a ftraight line, &c. Q. E. D. PROP. X. THEOR. See N. 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 fquare of the line made up of the half and the part produced. a II. T. Let the ftraight line AB be bisected in C, and produced to the point D; the fquares of AD, DB are 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 D draw b 31. 1. DF parallel to CE meeting EB cCor.39.1. produced in F, and through F draw FG parallel to AB meeting C • EC produced in G; and join The demonftration is the fame COR. Hence, because CD is half the fum of AD, DB, for it is equal to AC, DB together, and AC is half their difference AB; it is manifeft that the fquares of two lines AD, DB are together double of the fquares of CD half their fum, and AC half their difference. T PROP. XI. PROB. O divide a given ftraight line into two parts, fo that the rectangle contained by the whole, and one of the parts, fhall be equal to the fquare of the other part. Let ག Let AB be the given straight line; it is required to divide Book II it into two parts, fo that the rectangle contained by the whole, and one of the parts, fhall be equal to the square of the other part. Upon AB defcribe the fquare ABDC; bife&t AC in E, and join BE; produce CA to F, and make EF equal to EB; and upon AF defcribe a the fquare FGHA; AB is divided in H, so that the rectangle AB, BH is equal to the fquare of AH. d G a 46. 1. C I. 3. € 47. I. Produce GH to K: Because the ftraight line AC is bifccted in E, and produced to the point F, the rectangle CF, FA, together with the fquare of AE, is equal to the fquare of EF: d 6. 2. But EF is equal to EB; therefore the rectangle CF, FA, together with the fquare of AE, is equal to the fquare of EB: And the fquares of BA, AE are equal to the F fquare of EB, because the angle EAB is a right angle; therefore the rectangle CF, FA, together with the fquare of AE, is equal to the fquares of BA, AE: Take away the fquare of AE, which is common to both, therefore the remaining rectangle CF, FA is equal to the fquare of AB: And the figure FK is the rectangle con- E tained by CF, FA, for AF is equal to FG; and AD is the fquare of AB; therefore FK is equal to AD: Take away the common part AK, and the remainder FH C A K 11 B D is equal to the remainder HD: And HD is the rectangle contained by AB, BH, for AB is equal to BD; and FH is the square of AH: Therefore the rectangle AB, BH is equal to the fquare of AH: Wherefore the straight line AB is divided in H, so that the rectangle AB, BH is equal to the fquare of AH. Which was to be done. PROP. XII. THEOR. N obtufe angled triangles, if a perpendicular be IN obtufe ang any of the acute angles to the oppo fite fide produced, the fquare of the fide fubtending the obtufe angle, is greater than the fquares of the fides containing the obtufe angle, by twice the rectangle contained by the fide upon which, when produced, the perpendicular falls, and the ftraight line intercepted, without the triangle, between the perpendicular and the obtufe angle. Let BOOK II. Let ABC be an obtufe angled triangle, having the obtufe angle ACB, and from the point A let AD be drawn a perpendicular to a 12. 1. BC produced: The fquare of AB is greater than the squares of AC, CB, by twice the rectangle BC, CD. Because the ftraight line BD is divided into two parts in the c 4. 2. point C, the fquare of BD is equal to the fquares of BC, CD and twice the rectangle BC, CD: To each of thefe equals add the fquare of DA; and the fquares of BD, DA are equal to the fquares of BC, CD, DA, and twice the rectangle BC, CD: But 471 the fquare of BA is equal to the fquares of BD, DA, because the angle B at D is a right angle; and the fquare of CA is equal to the fquares of CD, DA: Therefore the fquare of BA is equal to the fquares of BC, CA, and twice the rectangle BC, CD; that is, the fquare of BA is greater than the fquares of BC, CA, by twice the rectangle BC, CD. Therefore, in obtufe angled triangles, &c. Q. E. D. See N. PROP. XIII. THEOR. I ing any of the c N every triangle, the fquare of the fide fubtending any of the acute angles, is lefs than the fquares of the fides containing that angle, by twice the rectangle contained by either of thefe fides, and the ftraight line intercepted between the perpendicular, let fall upon it from the oppofite angle, and the acute angle. Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the fides containing it, let fall the a 12. 1. perpendicular a AD from the oppofite angle: The square of AC, oppofite to the angle B, a A C D Because the straight line CB is divided into two parts in the 7.2. point D, or BD in C, the fquares of CB, BD are equal to b twice the rectangle contained by CB, BD, and the fquare of Book II. DC: To each of thefe equals add the fquare of AD; therefore the fquares of CB, BD, DA are equal to twice the rectangle CB, BD and the fquares of AD, DC: But the fquare of c AB is equal to the fquares of BD, DA, becaufe the angle c 47. 1. BDA is a right angle; and the fquare of AC is equal to the fquares of AD, DC: Therefore the fquares of CB, BA are equal to the fquare of AC and twice the rectangle CB, BD; that is, the fquare of AC alone is lefs than the fquares of CB, BA by twice the rectangle CB, BD. Wherefore, &c. Q. E. D. PROP. XIV. PROB. O defcribe a fquare that fhall be equal to a given Let A be the given rectilineal figure; it is required to defcribe a square that shall be equal to A. Defcribe the rectangular parallelogram BCDE equal to the a 45. I. rectilineal figure A. If then the fides of it BE, ED are equal to one another, it is a fquare, and what was required is now done: But if they are not equal, produce one of them BE to F, and make EF equal to ED, and bisect BF in G; and from the centre G, at the distance GB or c b 5. 2. GF, defcribe the femicircle BHF, and produce DE to H, and join GH: Therefore, because the straight line BF is divided into two equal parts in the point G, and into two unequal at E, the rectangle BE, EF, together with the fquare of EG, is equal to the fquare of GF: But GF is equal to GH; therefore the rectangle BE, EF, together with the fquare of EG, is equal to the square of GH: But the fquares of HE, EG are equal to c 47. I. the fquare of GH; therefore the rectangle BE, EF, together with the fquare of EG, is equal to the fquares of HE, EG: Take away the fquare of EG, which is common to both; and the remaining rectangle BE, EF is equal to the fquare of EH: But the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the fquare of EH; but BD is equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the fquare of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the fquare defcribed upon EH. EH. Which was to be done. |