Book II. PROP. IX. THEOR. F a straight line be divided into two equal, and also into 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. a II. I. с 5. 1. Let the straight 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 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 EA, EB; through D b 31. 1. draw b DF parallel to CE, and through F draw FG parallel to AB; and join AF: Then, because AC is equal to CE, the angle EAG is equal to the angle AEC; and because the angle ACE is a right angle, the two others AEC, EAC tod 32. 1. gether make one right angled; and they are equal to one an other; each of them therefore E half a right angle, and EGF a f 6. 1. c 29. 1. right angle, for it is equale to the interior and opposite 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 equalf 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 opposite 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 tof the fide DB. Now, because AC CE, ACCE2, and AC2+CE2AC2. But gAE-AC2+CE2; therefore AE22AC2. Again, because EG=GF, EGGF2, EG2+GF2=2GF2. But EF EG2+ GF2; therefore, EF2GF22CD2, becauseh CD= GF. And g 47. h 34: And it was shown that AE2AC2; therefore, AE2+EF Book II. 2AC2+2CD2. Butg AF AE2+EF2, and AD2+DF= AF2, or AD2+DB2 = AF2; therefore, also AD2+DB2= 2AC2+2CD2. Therefore, if a straight line, &c. Q. E. D. Otherwise: a "Because AD2 = AC2 + CD2 + 2AC.CD, and DB2 + a 4.2. “2BC.CD=b BC2+CD2=AC+CD2, by adding equals to b 7. 2. equals, AD2+DB2+2BC.CD=2AC2+2CD2+2AC.CD; " and therefore taking away the equal rectangles 2BC.CD and "2AC.CD, there remains AD2+DB22AC2+2CD2." PROP. Χ. THEOR. F 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. Let the straight line AB be bisected in C, and produced to the point D; the squares of AD, DB are double of the squares of AC, CD. From the point C draw a CE at right angles to AB, and a II. I. 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 straight line EF meets the parallels EC, FD, the angles CEF, EFD are equal to two right c 29. 1. angles; and therefore the angles BEF, EFD are less than two right angles: But straight lines, which with another straight line make the interior angles, upon the same side, less than two right angles, do meetd if produced far enough: Therefore d Cor. 29.1. EB, FD will meet, if produced towards B, D; let them meet in e 5. 1. Book II. 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 32. 1. half a right anglef: 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 g15. 1. 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; wherefore also the fide DB is equal to the fide DG. Again, because EGF is half a right h 6. r. E F angle, and the angle at i 34. 1. Fa right angle, being the fide GF is equalh to the G k fide FE. And because EC=CA, EC2+CA2CA2. Now * 47. 1. AE AC2+CE2; therefore, AEAC2. Again, because EF FG, EF FG2, and EF2+FG2EF2. But EG EF2+FG2; therefore, EG2=2EF2; and fince EF= CD, EG2CD2. And it was demonstrated, that AE 2AC2; therefore, AE2+EG2AC2+2CD2. Now, AG= AE2+EG2, wherefore AGAC2+2CD2. But AG2k AD2+DG2=AD2+DB2, because DG=DB: Therefore, AD2+DB22AC2+2CD2, Wherefore, if a straight line, &c. Q. E. D. PROP. Book II. Th PROP. XI. PRO В. O divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, may be equal to the square of the other part. 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. b10.1. C 3. 1. Upon AB describe a the square ABDC; bisect AC in E, a 46. Iand 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 bisected 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: d 6. 2. 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 equale to the square of EB, be- F G е 47. г. cause the angle EAB is a right to the remainder HD. But HD is the rectangle AB.BH for AB is equal to BD; and FH is the square of AH; there fore 1 Book II. fore the rectangle AB.BH is equal to the square of AH: Wherefore the ftraight line AB is divided in H so, that the rectangle AB.BH is equal to the square of AH. Which was to be done. N obtuse angled triangles, if a perpendicular be drawn from any of the acute angles to the oppofite fide produced, the square of the fide fubtending the obtuse angle is greater than the squares of the fides containing the obtuse angle, by twice the rectangle contained by the fide upon which, when produced, the perpendicular falls, and the straight line intercepted between the perpendicular and the obtuse angle. Let ABC be an obtufe angled triangle, having the obtufe a 12. 1. angle ACB, and from the point A let AD be drawna perpendicular to BC produced: The square of AB is greater than the squares 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, BD2 = b BC2 + CD2 + 2BC.CD; add AD2 to both: Then BD2+AD2 BC2+CD2+AD2 + C47. 2BC.CD. But ABBD2+AD2c, and AC CD2+AD2c; therefore, ABBC2+AC2+2BC.CD; that is, AB2 is greater than BC2+B C D AC2 by 2BC.CD. Therefore, in obtuse angled triangles, &c. Q. E. D. PROP. |