Book It. PROP. IX. THEOR. TF 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 11.1. 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 C 5. I. angle EAG is equal c 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 another; each of them therefore E T C D B half a right angle, and EGF a € 29. I. right angle, for it is equal e to the interior and opposite angle ECB, the remaining angle EFG is half a right angle; there fore the angle GEF is equal to the angle EFG, and the fide f 6. 1. EG equal to the side GF: Again, because the angle at B is half a right angle, and FDB a right angle, for it is equal e 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 lide DB. Now, be cause AC-CE, AC=CE?, and ACP+CE2=2ACP. But & 47. I. & AE=AC +CE; therefore AE2AC%. Again, because EG=GF,EG?=GF, EG+GF2=2GF2. But EF EG2+ h 34. 1: GF2; therefore, EF=2GF=2CD2, because h CD= GF. And And it was shown that AE2=2AC?; therefore, AE+EF? Book II. 2AC2+2CD2. But & AF AE+EF?, and ADP+DF = AF, or ADP+DBP = AF; therefore, also ADP+DB 2 AC-+2CD?. Therefore, if a straight line, &c. Q. E. D. Otherwise : “ Because ADP ACP + CD2 + 2AC.CD, and DBP + 2 4.2. 66 2BC.CD=) BCP+CD2ACP+CD2, by adding equals to b7. 2. “ equals, ADP+DB2+2BC.CD=2AC2+2CD2+2 AC.CD; " and therefore taking away the equal rectangles 2 BC.CD and * 2 AC.CD, there remains ADP+DB=2ACP+2CD.” 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 bifected, and of the square of the line made up of the half and the part produced. Let the straight line A B 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 11. 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 c 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 Book II. in G, and join AG: Then, because AC is equal to CE, the angle CEA is equale to the angle E AC; and the angle A CE e s. 1. is a right angle; therefore each of the angles CEA, EAC is $ 32. 1. half a right anglef : For the same 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 15. 1. & half a right angle, for they are vertically oppofite; but C 29. 1. BDG is a right angle, because it is equalc 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 also the side DB is equal h to the side DG. Again, because EGF is half a right E F angle, and the angle at F a right angle, being i 34. I. equali to the opposite an gle ECD, the remaining C с B В D "G fide FE. And because EC=CA, ECP+CA2=2CA?. Now k 47. 1. AEK ACP+CE?; therefore, AE LAC?. Again, be caufe EF=FG, EF=FG?, and EF?+FG=2EF. But EG=k EFP+FG?; therefore, EG=2EF?; and since EF= CD, EG’=2CD?. And it was demonstrated, that AE? 2AC2; therefore, AE+EG-2AC2+2CD?. Now, AG= AE?+EG?, wherefore AG-AC2+2CD? But AG?k= AD?+DG2 = ADP+DB, because DG = DB: Therefore, AD?+DB2=2ACP+2CD’, Wherefore, if a straight line, &c. Q. E. D. PROP Book II. PROP. XI. PROB. To 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, thall be equal to the square of the other part. Upon AB describe a the square ABDC; bisect AC in E, 2 46. Iand join BE; produce CA to F, and make c EF equal to 3.1: 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 АН. 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: 16. 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 € 47, 1. G equale to the square of EB, be. F cause the angle EAB is a right angle; therefore the rectangle CF.FA, together with the square of AE, is equal to the squares of BA, ІН AE: take away the square of A B AE, which is common to both, therefore the remaining rectangle CF.FA is equal to the square of AB. Now the figure FK is the E rectangle 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 c AK, and the remainder FH is equal K 'D 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 Book II. fore the rectangle AB.BH is equal to the square of AH: Wherefore the straight line AB is divided in À so, that the rectangle AB.BH is equal to the square of AH. Which was to be done. N obtufe angled triangles, if a perpendicular be drawn from any of the acute angles to the opposite 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 obtuse angled triangle, having the obtuse a 12. I. angle ACB, and from the point A let AD be drawn a per pendicular to BC produced : The square of AB is great- A divided into two parts in the b 4. 2. point C, BD? = BC? + CD? + BD2+AD=BC?+CD2 +AD? + c 47. I. 2BC.CD. But AB=BD2+ADPC, and AC=CD2+AD2c; there- D C PROP. |