*29.1. : +3.1. to AB, and make † it equal to AC or CB, and *31.1. 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 *12 Ax. 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, 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 angles CEA, EAC is half * a right angle. 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* half a right angle, for they are vertically opposite; but BDG is a right angle, because it is equal to the alternate angle DCE; therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; * 5.1. *32.1. *15. 1. *29.1. wherefore also the side *6. BD is equal to the side and that the angle at F * *34.1. it is equal to the opposite angle ECD, the remaining angle FEG is half a right angle, and therefore equal to the angle EGF; wherefore #6.1. also the side 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 *47.1. of CA: but the square of EA is equal to the squares of EC, 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, *47. 1. FE; therefore the square of EG is double of the square of EF; and EF is equal † to CD; where- +34. 1. fore 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: but the square of AG is equal to *47.1. 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 *47. 1. 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. * 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* *46. 1. AC in E, and join BE; produce CA to F, and *10. 1. make* EF equal to EB, and upon AF describe * -3.1. the square FGHA: AB shall be divided in H, *46. 1. so that the rectangle AB, BH is equal to the square of AH. Produce GH to K: and because the straight line AC is bisected in E, and produced to the point F, the rectangle CF, FÁ together with G *6.2. * the square of AE, is equal to the square of EF † Const. but EF is equal to EB; therefore the rectangle CF, FA, together with the square of AE, is equal to the square of EB: but the squares of BA, AE are F *47. 1. 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 E squares of BA, AE: take away the square of AE, which is common to both, therefore the re A C HB K D +3 Ax. maining rectangle CF, FA is equal to the square of AB: but the figure FK is the rect+30Def. angle contained by CF, FA, for AF is equal † to FG; and AD is the square of AB; therefore +1 Ax. FK is equal† to AD: take away the common +3. Ax. part AK, and the remainder FH is equal † to the remainder HD: but HD is the rectangle +30Def. 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 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 perpendicular to BC produced: *12. 1. the square of AB shall be 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; there- B fore the squares of BD, DA are * of *4.2. equal to the squares of BC, CD, DA, and twice †2 Ax. the rectangle BC, CD: but the square of BA is equal to the squares of BD, DA, because the *47. 1. angle at D is a right angle; and the square of CA is equal to the squares of CD, DA; there- *47. 1. fore the square of BA is equal to the squares 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. In every triangle, the square of the side subtending either 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* *12. 1. AD from the opposite angle: the square of AC, opposite to the angle B, shall be less than the 7.2. 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 to twice the rectangle contained by CB, BD, and thè square of DC; to each of these equals add the square of AD; therefore the squares of CB, BD, DA, are +2 Ax. equal to twice the rectangle CB, BD, and the squares of AD, DC: but the square of AB B *47.1. 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 ÂC, 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. Secondly, let AD fall witlıout the triangle ABC: then, because the angle at D is a right angle, the angle ACB is *16.1. greater than a right angle; and therefore the square of *12.2. AB is equal to the squares of *3.2. * B C AC, CB, and twice the rectangle BC, CD: to each of these equals add the square of BC; +2 Ax. therefore the squares of AB, BC are equal† to the square of AC, and twice the square of BC, and twice the rectangle BC, CD: but because BD is divided into two parts in C, the rectangle DB, BC is equal to the rectangle BC, CD and the square of BC; and the doubles of these are equal: therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC: therefore the square of AC alone is less than the squares of AB, BC, by twice the rectangle DB, BC. |