the rectangle AB, BC, is equal to the gnomon AOH: to each of these add XH, which is equal to the square of AC; therefore four times the rectangle AB, BC, together with the square of AC, is equal † to the gnomon AOH and the square XH: but the gnomon AOH and XH make up the figure AEFD, which is the square of AD: therefore four times the rectangle AB, BC, together with the square of AC, is equal to the square of AD, that is, of AB and BC added together in one straight line. Wherefore, if a straight line, &c. Q. E. D. 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. Let the straight line AB be divided into two equal parts at the point C, and into two unequal parts at the point D: the squares of AD, DB, shall together be double of the squares of AC, CD. * †1Ax. Cor. 4. 2. and 34. 1. +2 Ax. 11. 1. † * 3.1. 31. 1. 32.1. From the point C draw* CE at right angles to AB, and make it equal +_ to AC or CB, and join EA, EB; through D draw* DF parallel to CE, and through F draw FG parallel to BA; and join AF. Then, because AC is equal to CE, the angle EAC is equal to the *5.1. angle AEC: and because the angle ACE is a right angle, the two others AEC, EAC, together make one right angle; and they are equal to one another; therefore each of them 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 half a right angle, and EGFaright angle, for it is equal * 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 side 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 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 to the side ĎB. *6.1. A * * E F * 29. 1. CD B 6. 1. 29.1. 47. 1. † 47.1. * 34. 1. *47.1. * 11. 1. † 3.1. *31.1. * 29. 1. * 12 Ax. * * And because AC is equal to CE, the square of AC is equal to the square of ĈE; therefore the squares of AC, CE, are double of the square of AC: but the square of AE is equal to the squares of AC, CE because ACE is a right angle; therefore the square of AE 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 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 of GF; and GF is equal 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: but the square of AF is equal 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. * 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 shall be double of the squares of AC, CD. From the point C draw* CE at right angles to AB, and maket it equal to AC or CB, and join AE, EB; through E draw* 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 will meet* if produced far enough; therefore EB, FD will * * meet, if produced, towards B, D: let them meet in G, A E C B D 5. 1. 32.1. • 15. 1. 29. 1. 6.1. 34. 1. angle FEG is half a right angle, and therefore equal to the angle EGF; wherefore also the side GF is equal* *6. 1. 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, 47. 1. CA; therefore 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; 47. 1. therefore the square of EG is double of the square of EF; and EF is equal to CD; wherefore the square of † 34. 1. 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: but 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 47. 1. 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. 47. 1. E * 46. 1. * 10. 1. 3.1. * 46. 1. * 6.2. * 47.1. † 3 Ax. + 30 Def. † 1 Ax. + 3 Ax. † 30 Def. 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* the square ABDC; bisect* AC in E, and join BE: produce CA to F, and make* EF equal to EB, and upon AF describe* the square FGHA: AB shall be divided in H, so that the rectangle AB, BH is equal to the square of AH. F square Produce GH to K: and 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 to the square of EF: but EF is equal to EB; therefore the rectangle CF, FA, together with the of AE, is equal to the square of EB: but the squares of BA, AE, are equal* to the square of EB, because 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, AE: take away the square of AE, which is common to both, therefore the remaining rectangle CF, FA, is equal to the A E C HB K D square of AB: but the figure FK is the rectangle contained by 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 AK, and the remainder FH is equal to the remainder HD: but 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 either of the acute angles to the opposite side pro duced, 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* perpendicular to BC produced: the square of *12. 1. AB shall be greater than the squares of AC, CB by twice the rectangle BC, CD. A • 4.2. D Because the straight line BD is divided into two parts in the points 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; therefore the squares of B 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 angle; and the square of CA is 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. PROP. XIII. THEOR. +2 Ax. 47.1. 47.1. In every triangle, the square of the side subtending either See N. 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. 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 perpendicular* AD from the op- • 12. 1. posite angle: the square of AC, opposite to the angle B, shall be 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 |