Book II. PROP. IX. THEOR. DF parale AC or CB, chat rightares of Acuares of If 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. 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 drawa CE at right angles to AB, and •11. 1. make it equal to AC or CB, and join EA, EB; through D. draw bDF parallel to CE, and through F draw FG parallel b31. 1. to AB; and join AF: Then, because AC is equal to CE, the angle EAC is equal to the angle AEC; and because the c5. 1. angle ACE is a right angle, the two others AEC, EAC together make one right angled; and they are equal to one • 32. 1. another; each of them therefore is half of a right angle. For the same reason 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 – C D B . half a right angle, and EGFZ A right angle, for it is equale to the interior and opposite an- 29. 1. gle ECB, the remaining angle EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the side EG equalf to the side GF: Again, because the f6. 1. 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 side DF tof the side DB :. And because 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 equals to the squares of AC, 547. I. CE, because ACE is a right angle; therefore the square of EA is double of the square of AC: Again, because 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 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 b 34. 1. of the square GF; and GF is equalh 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 . i 47. 1. of AC, CD: And the square of AF is equali 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. :: PROP. X. THEOR. 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. 'au. i. From the point C draw a CE at right angles to AB: And make it equal to AC or CB, and join AE, EB; 31. 1. through E draw bEF parallel to AB, and through D draw DF parallel to CE: And because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are c 29. 1. equal to two right 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 - 12 Ax. upon the same side less than two right angles, do meetd if produced far enough: Therefore EB, FD shall meet, if pro duced towards B, D: Let them ineet in G, and join AG: 5. 1. Then, because AC is equal to CE, the angle CEA is equale to the angle EAC; and the angle ACE is a right angle ; f32. 1. therefore each of the angles CEA, EAC is half a right angle. For the same reason, each of the angles CEB, EBC is half Book II. a right angle; therefore AEB is a right angle: And because EBC is half a right angle, DBG is alsof half a right f15. 1. angle, for they are vertically opposite ; but BDG is a right angle, because it is equale to the alternate angle DCE; 29. 1. therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore also the side BD is equals to 66. 1. the side DG. Again, because EGF is half a right angle, and that the angle at F is a right angle, because it is equalb 34. 1. to the opposite angle A ECD, the remaining angle FEG is half a right angle, and equal to the angle EGF; wherefore also the side GF is equals to the side FE. And because EC is equal to CA, the square of EC is equal to the square of CĂ; therefore the squares of EC, CA are double of the square of CA: But the square of EA is equali to the squares of EC, CA; there- i 47. 1. fore 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 equali 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 square of AC, CD: And the square of AG is equali to the squares of AE, EG; therefore the square of AG is double of the squares of AC, CD: But the squares of AD, GD 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. ed to the the square of EG, that the square the EG is doverefore the demonsac; there AC, CDEG; there: 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, shall 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. ? 46. 1. Upon AB describea the square ABDC; bisectb AC in 10.1. E, and join BE; produce CA to F, and makec Ef equal € 3. 1 " to EB, and upon AF describea the square of 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 bi sected in E, and produced to the point F, the rectangle 26. . CF, FA, together with the square of AE, is equald 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 EB, because the angle EAB is a right ІН В K D part AK, and the remainder FH 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 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. XII. THEOR. In obtuse angled triangles, if a perpendicular be drawn from any of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle. Let ABC be an obtuse angled triangle, having the obtuse angle ACB, and from the point A let AD be drawna • 12. 1. perpendicular to BC produced: The square of AB is greater than the squares of AC, CB, by twice the rectangle BC, CD. Because the straight line BD is divided into two parts in the point C, the square of BD is equal bto the squares of BC, CD, 4. 2. and twice the rectangle BC, CD: To each of these equals add the square of DA; and the squares of BD, DA, are equal to the squares of BC, CD, DA, and twice the rectangle BC, CD: But the square of BA is equal B D • 47. 1. to the squares of BD, DA, be. cause the angle at D is a right angle ; and the square of CA is equal to the squares of CĎ, DA: Therefore the square of BA is equal to the squares of BC, CA, and twice the rectangle BC, CD; that is, the square of BA is greater than the squares of BC, CA, by twice the rectangle BC, CD. Therefore, in obtuse angled triangles, &c. Q. E. D. |