line EF falls upon the parallels EC, FD, the angles CEF, EFD, are equal to two right angles. Therefore the 29. 1. angles FEB, EFD, are less than two right angles. But right lines making with a third line angles together less than two right angles being infinitely produced will meet. Wherefore EB, FD, produced, will meet to- 12 Ax. wards the parts B, D. Let them be produced and meet in the point G, and let AG be drawn. Then because AC is equal to CE, and the angle AEC shall be equal to the angle EAC. But the angle at c is a right one; therefore the angle EAC or AEC is half a right one. By the same reasoning the angle CEB or EBC E C B F G But BDG e 5. 1. square h 47. 1. is half of a right one. Wherefore AEB is a right angle, and since EBC is half a right angle, DBG will also be ƒ 15. 1. half a right angle, since it is vertical to CBE. is a right angle also, for it is equal to the alternate angle DCE. Wherefore the angle DBG is equal to the angle DGB. And thence in the triangle DBG the sides BD, DG, are equal. Again, because the right lines BD, EF, are parallel, and the right line EG falls upon them, the angle DBG will be equal to the angle GEF, and in the same manner GBD will be equal to EGD. Wherefore the angles GEF, EGF, are equal, and in the triangle FGE the side GF is equals to the side EF. And g 6. 1. since Ec is equal to CA, and the square of EC equal to the square CA, therefore the squares of EC, CA, together, are double of the square ca. But the of EA is equal" to the of EC, CA, and squares quently double of the square of ca. Again, because EF is equal to GF, the square of GF also is equal to the square of FE. Wherefore the squares of GF, FE, are double to the square But the square of EG ish equal to the squares of GF, FE. Therefore the square of EG is double to the square of EF, but EF is equal to CD. Wherefore the square of EG shall be double to the square of CD, But it was proved 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, GE. 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 right line, &c. Q. E. d. of FE. *Cor. 4.2. † 4.2. a 46.1. b 6.2. с © 47. 1. The same by Algebra. Put a for the line AB, and b for the added line BD, then shall b2 + a2 + 2 ab + b2 = 1 a2 + ¦ a2 +2ab+ 2 2 b2. For 2. a2* = 1 a2, and 2. a + b † = ¦ a3 +2b2+2 a b. Q. E. D. PROPOSITION XI. PROBLEM. To cut a given right line into two parts so that the rectangle which is contained under the whole line and one part may be equal to the square made of the other part. Let AB be a given right line. It is required to cut it so that the rectangle contained under the whole line and one part may be equal to the square made of the other part. Let the square ABCD be describeda on F of EB, E C H K. G B D For since the right line AC is bisected in E, and AF is added directly thereto, the rectangle under CF, FA, together with the square of AE, shall be equal to the square of EF; but EF is equal to EB; therefore the rectangle under CF, FA, together with the square of AE, is equal to the square made on EB. But the squares of AB, AE, are equal to the square for the angle at A is a right one: therefore the rectangle under CF, FA, together with the square of AE, is equal to the squares of BA, AE; and if the square of AE, which is common, be taken away, the remaining rectangle under CF, FA, is equal to the square of AB. But FK is the rectangle under CF, FA; for AF is equal FG, and the square of AB is AD. Therefore the rectangle FK is equal to the square of AD. And if AH, which is common, be taken away, therefore the remaining FH is equal to the remaining But HD is the rectangle under AB, BH, for AB is equal to BD, and FH is the square of AH. Therefore the rectangle AB, BH, shall be equal to the square of And so the given right line AB is cut in н, so HD. that the rectangle under AB, BH, is equal to the Which was to be done. of AH. PROPOSITION XII. square THEOREM. In obtuse angled triangles, the square of the side subtending the obtuse angle is greater than the squares, which are made by the sides containing the obtuse angle by twice the rectangle contained under one of the sides, which are about the obtuse angle, viz. that on which produced the perpendicular falls, and the line taken without between the perpendicular and obtuse angle. B A D Let ABC be an obtuse angled triangle, having the obtuse angle BAC, and draw from the point в the perpendicular BD to CA produced. The square of BC is greater than the squares of BA, AC, by twice the rectangle which is contained under CA, AD. For because the right line CD is any how cut in the point A, the square of CD shall be equal to the squares of CA, AD, and to twice the rectangle AC, AD. To each of these equals add the square of DB. Therefore the squares of CD, DB, are equal to the C squares of CA, AD, and twice the rectangle CA, AD. But the square of CB is equal to the squares of CD, DB, for the angle at D is a right one, since BD is perpendicular, and the square of AB is equal to the squares of AD, DB. Therefore the square of CB is equal to the squares of CA, AB, and twice the rectangle under CA, Therefore the square of CB is greater than the squares of CA and AB by twice the rectangle contained under CA, AD. Therefore in obtuse angled triangles, &c. Q. E. D. AD. b The same by Algebra. ь Put a = CB, b = ca, c = AB, d = DB, and e = a 4. 2. b 47.1. * 47.1. † 4. 2. a 12. 1. b 7. 2. c 47. 1. a 7. 2. PROPOSITION XIII. THEOREM. In acute angled triangles, the square of the side subtending any of the acute angles is less than the squares of the sides containing the acute angle, by twice a rectangle under one of the sides about the acute angle, viz. on which the perpendicular falls, and the line assumed within the triangle from the perpendicular to the acute angle. B D Let ABC be any acute angled triangle having an acute angle at B, and from the point a drawa AD perpendicular to BC. The square of AC is less than the squares of AB, BC, by twice the rectangle contained under BD, BC. For since the right line BC is any how cut in the point D, the squares of BD, BC, shall be equalb to twice a rectangle under CB, BD, together with the square of DC. And if the square of AD be added to both, then the squares of CB, BD, and DA, are equal to twice the rectangle under CB and BD, together with the squares of AD and DC. But the square of AB is equal to the squares of BD, DA, for the angle at D is a right one. And the square of AC is equal to the squares of AD, DC. Therefore the squares of CB and BA are equal to the square of AC together with twice the rectangle under CB and BD. Wherefore the square of AC only is less than the squares of CB and BA, by twice the rectangle under CB and BD. Therefore in acute angled triangles, &c. Q. E. D. This proposition will hold true in obtuse and right angled triangles as well as acute, as may be perceived by the following demonstration. For since Let ABC be an obtuse angled triangle, and the perpendicular fall without the triangle, as AD. BD is divided into two parts in the point c, the squares of BC, BD, are equal to twice the rectangle of BC, BD, together with the square of DC. And if to each of these equals there be added the square of AD, the squares D of CB, BD, and AD, will be equal to twice the rectangle of BC, BD, together with the sum of the squares of AD, DC. But the squares of BD, AD, are equal to the square .. 47. 1. of square of AC; AB, and the of AD, DC, to the squares whence the square of AC is less than the sum of the squares of BC, BA, by twice the rectangle BC, BD. Again, if the side AC be perpendicular to BC, then is BC the right line between the perpendicular and the acute angle at B; and it is manifest that the squares of AB, BC, are equal to the square of AC and twice the square of BC. B The same by Algebra. BC, b = AB, c = AC, d = BD, e = DC, Then ce + ƒ2* but c2d2 = 2 c d + e2 + To describe a square equal to a given right lined figure. Let A be the given right lined figure. It is required to describe a square equal thereto. a But a Describe the right angled parallelogram BCDE equal a 45. 1. to the right lined figure A. Then if BE is equal to ED, what was proposed will be done, for the square BD is described equal to the given right lined figure A. if BE, ED, are unequal, produce one of them to F, and make EF equal to ED. Then BF being bisected in G, about which as a centre with the distance GB or GF, describe the semicircle BHF, and let DE be produced to H, and draw GH. Now because the right line BF is bisected at G, and divided into two unequal parts in E, the rectangle under BE, EF, together with the square of EG, will be equal to the square of GF. But GF is equal to GH. Wherefore the rectangle under BE, EF, together with the square of GE, is equal to the square of GH. But the of A b 5.2. square of GH € 47. 1. HE, GE, are equal to the squares therefore the rectangle under BE, EF, together with the |