Book II. a 11. I. b 31. 1. C 5. 1. d32. 1. PROP. IX. THEOR. Fa ftraight line be divided into two equal, and alfo 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 squares of AD, DB are together double of the fquares 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 EA, EB; through D draw b DF parallel to CE, and through F draw FG parallel to AB; and join AF: Then, because AC is equal to ĈE, 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 angled; 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: G E And because the angle GEF is A C D B e 29. 1. right angle, for it is equale to the interior and 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 f 6. 1. EG equal to the fide GF: Again, because the angle at B is half a right angle, and FDB a right angle, for it is equale 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. Now, becaufe AC CE, AC=CE', and AC+CE22AC2. But 847. AE AC+CE; therefore AE2AC2. Again, because EG=GF,EG2=GF2, EG2+GF2=2GF2. But ĔF2 EG2+ h 34. GF2; therefore, EF2GF2=2CD2, because h CD GF. And And it was shown that AE2AC2; therefore, AE+EF Book II. 2AC2+2CD2. But g AF AE2+EF2, and AD2+DFAF2, or AD2+DB2 = AF2; therefore, alfo AD2+DB2= 2AC2+2CD2. Therefore, if a straight line, &c. Q. E. D. Otherwise : "Because AD2 AC2+ CD2+2AC.CD, and DB2+ a 4.2. "2BC.CDb BC2+CD2 AC4+CD2, by adding equals to b 7. 2. "equals, AD+DB2+2BC.CD = 2AC2+2CD2+2AC.CD; "and therefore taking away the equal rectangles 2BC.CD and "2AC.CD, there remains AD2+DB2 2AC2+2CD2." PROP. X. THEOR. F a 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 bifected, and of the fquare 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 squares of AC, CD. 29. 1.) From the point C draw a CE at right angles to AB, and a 11. 1. make it equal to AC or CB; join AE, EB; through E draw b EF parallel to AB, and through D draw DF parallel b 31. 1. 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 less than two right angles: But ftraight lines, which with another straight line make the interior angles, upon the fame fide, less than two right angles, do meet d if produced far enough: Therefore d Cor. 29.1. EB, FD will meet, if produced towards B, D, let them meet in 5. I. Book II. in G, and join AG: Then, because AC is equal to CE, the angle CEA is equale to the angle EAC; and the angle ACE is a right angle; therefore each of the angles CEA, EAC is 32. 1. half a right angle: For the fame reason, each of the angles CEB, EBC is half a right angle; therefore AEB is a right angle: And because EBC is half a right angle, DBG is also g half a right angle, for they are vertically oppofite; but c 29. 1. BDG is a right angle, because it is equal to the alternate angle DCE; therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; whereh 6. 1. fore alfo the fide DB is equal to the fide DG. Again, becaufe EGF is half a right g 15. I. angle, and the angle at i 34. 1. equal to the oppofite an- the fide GF is equal to the E F B G fide FE. And because EC CA, EC2+CA2=2CA2. Now k 47. 1. AE AC2+CE; therefore, AEAC2. Again, because EF FG, EF FG2, and EF2+FG=2EF2. But EG2k EF2+FG2; therefore, EG2EF2; and fince EF= CD, EG2 2CD2. And it was demonftrated, that AE2; 2AC2; therefore, AE2+EGAC2+2CD2. Now, AG AE+EG2, wherefore AGAC2+2CD2. But AGk= AD2+DG2 = AD2+DB2, because DGDB: Therefore, AD2+DB2=2AC2+2CD2, Wherefore, if a ftraight line, &c. Q.E. D. PROP. PROP. XI. PROB. Book II. T O divide a given straight line into two parts, fo that the rectangle contained by the whole, and one of the parts, may be equal to the fquare of the other part. Let AB be the given straight line; it is required to divide it into two parts, fo that the rectangle contained by the whole, and one of the parts, shall be equal to the fquare of the other part. C 3. 1. Upon AB defcribe a the fquare ABDC; bifect AC in E, a 46. r and join BE; produce CA to F, and makec 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 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 fquare of AE, is equal d 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 E G H B K D to the remainder HD. But HD is the rectangle AB.BH for AB is equal to BD; and FH is the fquare of AH; there fore € 47. r. Book II. fore the rectangle AB.BH is equal to the fquare of AH: Wherefore the ftraight line AB is divided in Ĥ fo, that the rectangle AB.BH is equal to the fquare of AH. Which was to be done. IN N obtufe angled triangles, if a perpendicular be drawn from any of the acute angles to the oppofite 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 between the perpendicular and the obtufe angle. Let ABC be an obtufe angled triangle, having the obtuse a 12. 1. angle ACB, and from the point A let AD be drawn a perpendicular to BC produced: The fquare of AB is greater than the fquares of AC, CB, by twice the rectangle BC.CD. b 4. 2. Because the straight line BD is divided into two parts in the point C, BD2b BC2 + CD2 + 2BC.CD; add AD2 to both: Then BD2+AD2 BC2+CD2+AD2 + C47. I. 2BC.CD. But AB BD2+AD2c, and AC CD2+AD2 c; therefore, AB2 BC2+AC2+2BC.CD; that is, AB2 is greater than BC2+B C A AC2 by 2BC.CD. Therefore, in obtufe angled triangles, &c. Q. E. D. PROP. |