Book II. PROP. IX. THEOR, 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. G 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 sqnares 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 EX, EB; through D draw > DF parallel to CE, and through F draw FG parallel to AB; b 31. t. and join AF: then, because AC is equal to CE, the angle EAC is equal to the angle AEC; and because the angle c 5, 1. ACE is a right angle, the two others, AEC, EAC together make one right angled; and they are equal to one another; d 32. 1. cach of them therefore is half E of a right angle. For the same reason each of the angles CEB, EBC is half a right angle; and F 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 A с D B equale to the interior and oppo e 29. 1. site angle ECB, the remaining angle EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the side EG equal to the side GF: again, because the angle at f 6. 1 Bis half a right angle, and FDB half 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 side DF tof the side DB: and be. cause 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, CE, because ACE is a right angle ; therefore & 47. . 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 FG, 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 of the square h 34. 1. 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 AF, EF i 47.1. are double of the squares 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. If 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. a 11. 1. b 31. 1. c 29. 1. 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 drawa CE at right angles to A B: and make it equal to AC or CB, and join AE, EB; through E drawb EF 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 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 upon the same side less than two right angles, do meet d if produced far enough: therefore EB, FD shall meet, if produced towards B, D: let them meet 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 half a right anglef: for the same reason d 12. Ax. e 5. 1. 132.1. each of the angles CEB, EBC is half a right angle; therefore Book II. AEB is a right angle: and because EBC is half a right angle, w DBG is also half a right angle, for they are vertically oppo- f 15. 1. site; but BDG is a right angle, because it is equal to the al-c 29. 1. ternate angle DCE; therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore also the side BD is equal 6 to the side DG: again, g 6. 1. because EGF is half a E F right angle, and that the angle at F is a right angle, because it is equalh to the opposite angle ECD, the с B h 34. 1. remaining angle FEG is half a right angle, and A D equal to the angle EGF; wherefore also the side G GF is equal 5 to the side FE. And because EC is equal to CA, the square of EC is equal to the square of CA; therefore the squares of EC, CA are double of the square of CA: but the square of EA is equal i to the squares of EC, CA; therefore the square of EA ; 47. 1. 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 equal i 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 squares 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 equal i 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: Book IL PROP. XI.:: PROB. 1 TO divide a given straight line into two parts, so that the rectangle contained by the whole and one of the other parts shall be equal to the square of the other part. a 46. 1. bi). 1. C 3.1. d 6.2. e 47. 1. 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. Upon AB describe a the square ABDC; bisectb AC in E, and join BE ; produce CA to F, and make Ef equal to EB; and upon AF describe & 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 G H B E AD is the square of AB; therefore К. D and HD is the rectangle contained by AB, BH, for AB is cqual to BD; and FH is the square of AH: therefore the Tectangle 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. Book it, 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 drawn a perpen- & 12. 1. dicular 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 to the squares of BC, CD, and A 24.2. 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< to the squares of c 47.1. BD, DA, because the angle at D B с D is a right angle; and the square of CA is equals to the squares of CD, 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. |