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 equal h 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 of AC, CD: and the square of AF is i 47. 1. equal i 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. a 11. 1. b 31. 1. c 29. 1. 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. 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. c From the point C draw a CE at right angles to AB: and make it equal to AC or CB, and join AE, EB; through E draw b 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 d 12. Ax. 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 equal to the angle EAC; and the angle ACE is a right angle; therefore each of the f32. 1. angles CEA, EAC is half a right angle f: for the same reason, e 5. 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, ~ 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 to the side DG: again, g 6. 1. because EGF is half a right angle, and that the angle at F is a right angle, because it is equal to the opposite angle ECD, the remain E F ing angle FEG is half a right angle, and equal to the angle EGF; wherefore also the side i G GF is equal 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 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 equal 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 equal 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 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 II. a 46. 1. 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. Upon AB describe a the square ABDC; bisect AC in E, and b 10. 1. 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. c 3. 1. d 6. 2. 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: 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 e 47. 1. squares of BA, AE are equal to the A E F G H B square of EB, because the angle EAB C K D Book II. 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 perpendi- a 12. 1. cular 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 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 BD, DA, because the angle at D is a right B angle; and the square of CA is C A b 4.2. c 47. 1. D equal 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. Book II. See N. PROP. XIII. THEOR. IN every triangle, the square of the side subtending any of the acute angles is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle. of Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the a 12. 1. perpendicular a AD from the opposite angle: the square AC, opposite to the angle B, is less than the squares of CB, BA, by twice the rectangle CB, BD. First, Let AD fall within the triangle ABC; and because the straight line CB is divided into two parts in the point D, the squares of CB, BD are equal b 7. 12. b to twice the rectangle contain ed by CB, BD, and the square B A D C € 47. 1. but the square of AB is equal to the squares of BD, DA, because the angle BDA is a right angle; and the square of AC is equal to the squares of AD, DC: therefore the squares of CB, BA are equal to the square of AC, and twice the rectangle CB, BD, that is, the square of AC alone is less than the squares of CB, BA by twice the rectangle CB, BD. d 16. 1. e 12. 2. Secondly, Let AD fall without the triangle ABC: then, because the angle at D is a right angle, the angle ACB is greater than a right angle; and therefore the square of AB is equale to the squares of AC, CB, and twice the rectangle BC, CD: to these equals add the square of BC, and the B C D |