C 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 alfo f half a right angle, for they are vertically oppof 15. I. fite; but BDG is a right angle, because it is equal to the al- c 29. I. ternate angle DCE; therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore alfo the fide BD is equal to the fide 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 h to the oppofite angle ECD, the remaining angle FEG is half a right angle, and equal to the angle EGF; wherefore alfo the fide A GF is equals to the fide FE. And because EC is equal to CA, the fquare of EC is equal to the fquare of CA; therefore the fquares of EC, CA are double of the fquare of CA: But the fquare of EA is equal i to the fquares EC, CA; there-i 47. x. fore the square of EA is double of the square of AC: Again, becaufe GF is equal to FE, the fquare of GF is equal to the square of FE; and therefore the fquares of GF, FÈ are double of the fquare of EF: But the fquare of EG is equal i to the fquares of GF, FE; therefore the square of EG is double of the fquare of EF: And EF is equal to CD; wherefore the fquare of EGis double of the fquare of CD: But it was demonftrated, that the square of EA is double of the square of AC; therefore the squares of AE, EG are double of the fquares of AC, CD: And the square of AG is equal to the fquares of AE, EG; therefore the fquare of AG is double of the fquares of AC, CD: But the fquares of AD, GD are equal i to the fquare of AG: therefore the fquares of AD, DG are do ble of the fquares of AC, CD: But DG is equal to DB; therefore the fquares of AD, DB are double of the fquares of AC, CD: Wherefore, if a ftraight line, &c. Q. E. D. PROP. Book II. a 46. I. b 10. 1. C3. I. d 6.2. € 47. I. T PROP. XI. PROB. 'O divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, fhall 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, fhall be equal to the fquare of the other part. Upon AB defcribe a the fquare ABDC; bifect b AC in E, and join BE; produce CA to F, and make © EF equal to EB, and upon AF defcribe a the fquare FGHA; AB is divided in H, fo that the rectangle AB, BH is equal to the fquare of AH. G HB Produce GH to K: because the straight line AC is bifected 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 fquare of AE, is equal to the fquare of EB: And the fquares of BA, AE are equal to the F fquare of EB, because the angle EAB is a right angle; therefore the rectangle CF, FA together with the fquare of AE, is equal to the fquares of BA, AE: Take away the fquare of AE, A which is common to both, therefore the remaining rectangle CF, FA is equal to the fquare of AB: and the fi- E gure FK is the rectangle contained by CF, FA, for AF is equal to FG; and AD is the fquare of AB; therefore FK is equal to AD: Take away the common part AK, and the remainder FH is equal to the remainder HD : And HD is the rectangle contained by AB, BH, for AB is equal to BD; and FH is the fquare of AH. Therefore the rectangle AB, BH is equal to the fquare of AH: Wherefore the ftraight line AB is divided in H fo, that the rectangle AB, BH is equal to the fquare of AH. Which was to be done. C K D PROP. Book II. IN PROP. XII. THE OR. N obtufe angled triangles, if a perpendicular be drawn from any of the acute angles to the oppofite fide produced, the fquare of the fide fubtending the obtufe angle is greater than the fquares of the fides containing the obtufe angle, by twice the rectangle contained by the fide upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtufe angle. Let ABC be an obtufe angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn a perpen- a 12. I. dicular to BC produced: The fquare of AB is greater than the squares of AC, CB by twice the rectangle BC, CD. A b 4. 2. Because the ftraight line BD is divided into two parts in the point C, the fquare of BD is equal b to the fquares of BC, CD, and twice the rectangle BC, CD: To each of thefe equals add the fquare of DA; and the fquares of DB, DA are equal to the fquares of BC, CD, DA, and twice the rectangle BC, CD: But the fquare of BA is equal to the fquares of BD, DA, because the angle at D is a right B angle; and the fquare of CA is e C 47. I. C D qual to the fquares of CD, DA: Therefore the square of BA is equal to the fquares of BC, CA, and twice the the rectangle BC, CD; that is, the fquare of BA is greater than the fquares of BC, CA, by twice the rectangle BC, CD. Therefore, in obtufe angled triangles, &c. Q. E. D. 1 PROP. Book II: See N. a 12. I. b 4.2. € 47. I. d 16. 1. € 12. 2. PRO P. XIII. THE OR. every triangle, the fquare of the fide fubtending any of the acute angles, is lefs than the fquares of the fides containing that angle, by twice the rectangle contained by either of thefe fides, and the ftraight line intercepted between the perpendicular let fall upon it from the oppofite 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 fides containing it, let fall the perpendicular a AD from the oppofite angle: The fquare of AC, oppofite to the angle B, is lefs than the squares of CB, BA by twice the rectangle CB, BD. A First, Let AD fall within the triangle ABC; and because the ftraight line CB is divided into two parts in the point D, the fquares of CB, BD are aqual b to twice the rectangle contained by CB,BD, and the square of DC: To each of these equals add the fquare of AD; therefore the fquares of CB, BD, DA, are equal to twice the rectangle CB, BD, and the fquares of AD, DC: But the fquare of AB is equal B D C to the fquares BD, DA, because the angle BDA is a right angle; and the fquare of AC is equal to the fquares of AD, DC: Therefore the fquares of CB, BA are equal to the square of AC, and twice the rectangle CB, BD, that is, the fquare of AC alone is lefs than the fquares of CB, BA by twice the rectangle CB, BD. Secondly, Let AD fall with- A C D fquares fquares of AB, BC are equal to the fquare of AC, and twice Book II. the fquare of BC, and twice the rectangle BC, CD: But becaufe BD is divided into two parts in C, the rectangle DB, BC is equal f to the rectangle BC, CD and the fquare of BC: And f 3. 2. the doubles of these are equal: Therefore the fquares of AB, BC are equal to the fquare of AC, and twice the rectangle DB, BC: Therefore the fquare of AC alone is less than the squares of AB, BC by twice the rectangle DB, BC. Laftly, let the fide AC be perpendicular to BC; then is BC the ftraight line between the perpendicular and the acute angle at B; and it is manifeft that the fquares of AB, BC are equal g to the fquare of AC and twice the fquare of BC: Therefore, in every triangle, &c. Q E. D. A g 47. I. O defcribe a fquare that shall be equal to a given See N. Trectilineal figure. Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A. Defcribe a the rectangular parallelogram BCDE equal to the a 45. §. rectilineal figure A. If then the fides of it BE, ED are equal to from the centre G, at the distance GB, or GF, defcribe the femicircle BHF, and produce DE to H, and join GH: Therefore because the straight line BF is divided into two equal parts in the point G, and into two unequal at E, the rectangle BE, EF, together with the square of EG, is equal to the fquare of b s. 2. GF: But GF is equal to GH; therefore the rectangle BE, EF, |