PRO P. VI. Plate I. Fig. 53. If a right-line A be divided into two equal parts, and another right-line E, added to the fame directly in one right-line, then the rectangle comprehended under the whole and the line added, (viz. A+E,) and the line added E, together with the Square which is made of the line A, is equal to the fquare of A+E taken as one line. I 4 I 2 I fay that Aq. ( (a) Q. & A) +AE+Eq=Q. A+E. (a) For, Q. ATE=Aq+Eq AE. Which was to be demonftrated. Coroll. Hence it follows, that if 3 right-lines E, EA, E† A be in arithmetical proportion, then the rectangle contained under the extreme terms E, EA, together with the fquare of the difference A, is equal to the square of the middle term E+A. PROP. VII. Fig. 50. If a right-line Z be divided any wife into two parts, the Square of the whole line Z, together with fquare made of one of the fegments E, is equal to a double rectangle comprehended under the whole line Z, and the faid fegment E, together with the fquare made of the other fegment A. I fay, that Zq+Eq=2ZE+Aq. For Zq (a) Aq+ Eq-1-2AE, and 2 ZE (6)—2 Eq+2AE. Which avas to be demonfirated. Coroll. Hence it follows, that the fquare of the difference of any two lines Z, E, is equal to the fquare of both the lines lefs by a double rectangle comprehended under the faid lines. (c) For Zq+Eq-2ZE—Aq—Q. Z-E. PRO P. VIII. Fig. 50. If a right-line Z be divided any wife into two parts, the rectangle comprehended under the whole line Z, and one of the fegments E four times, together with the square of the other fegment A, is equal to the fquare of the whole line Z, and the fegment E, taken as one line ZE. I fay, that 4ZE+Aq=QZ +E. For 2 ZE (0)= Zq+Eq Aq. Therefore 4 ZE + AqZq+Eq +2ZE (6)=QZE. Which was to be demonftrated. a4 & 3. Cor. 4. 2. a 4. 2. b 3.2. C 7. 2. and 3. ax. a 7. 2. and 3. ax. a 4.2. b byp. a 4, 2. If a right-line AB be divided into equal parts AC, CB and into unequal parts AD, DB, then are the fquares of the unequal parts AD, DB, together, double to the fquare of the half line AC, and to the fquare of the difference CD. I fay that ADq+DBq=2 ÁCq +2CDq. For ADq +DBq (a) = ACq + CDq + 2 ACD+DBq. But 2ACD (b) (2BCD)+DBq () CBq (ACq) + CDq. (d) Therefore ADq + DBq 2 ACq+2CDq. Which was to be demonftrated. This may be otherwise delivered and more eafily demonstrated thus; the aggregate of the fquares made of the fum and the difference of two right-lines A, E, is equal to the double of the squares made from thofe lines. For QA+E (a) Aq+ Eq+2AE, and Q. Ab cor. 7. 2. E(6)=Aq+Eq-2AE. These added tegether make 2Aq +2Eq. Which was to be demonftrated. a 4. 2. a 46. 1. b 10. 1. PROP. X. Fig. 53. If a right-line A be divided into trvo equal parts, and another line be added in a right-line with the fame, then is the Square of the whole line together with the added line (as being one line) together with the fquare of the added line E, double to the fquare of half Ä, and the added line E, taken as one line. I fay that Eq+Q.A + E, ¿. e. (a) Aq† 2 Eq42AE = 2 Q2 = A + 2 QAE, For 2 QA (6) Aq. And 2 QA+E (c) = 1⁄2 Aq + 2 Eq + 2AE. Which was to be demonftrated, PROP. XI. Fig. 55. To cut a right-line given AB, in a point G, so that the rectangle comprehended under the whole line AB, and one of the fegments BG, shall be equal to the Square, that is made of the other Segment AG. Upon AB (a) describe the square AC (b) Bifect the fide AD in E, and draw the line EB; from the line EA produced take EFEB. On AF make the square AH. Then is AH AB× BG. For For HG being drawn out to I; the rectangle DH-+EAq (c)=EFq (d) —EBq (e) —BAq+EAq: Thereforec 6. 2. is DH (ƒ)=BAq to the fquare AC. Take away AI d conftr. common to both, there remains the fquare AH=GC, € 47. 1. that is, AGqABX BG. Which was to be done. £ 3. ax. Schol. This propofition cannot be performed by numbers; * for there is no number that can be fo divided, that the product of the whole into one part fhall be equal to the fquare of the other part. PROP. XII. Plate I. Fig. 56. In obtufe-angled triangles as ABC, the fquare that is made of the fide AC, fubtending the obtufe angle ABC, is greater than the fquares of the fides BC, AB, that contain the obtufe angle ABC, by a double rectangle contained under one of the fides BC, which are about the obtufe angle ABC, on which fide produced the perpendicular AD falls, and under the Line BD, taken without the triangle from the point on which the perpendicular AD falls to the obtufe angle ABC. I fay that ACq=CBq+ABq+2CBX BD. For these are all equal ACq. (a) CDq+ADq. (b) CBq+2CBD+BDq+ADq; Scholium. Hence, the fides of any obtufe-angled triangle ABC being known, the fegment BD intercepted betwixt the perpendicular AD, and the obtuse angle ABC, as also the perpendicular it felf AD, shall be easily found out. Thus, Let AC be 10, AB 7, CB 5. Then is ACq 100, ABq 49, CBq 25. And ABq+CBq=74. Take that out of 100, then will 26 remain for 2 CBD. Wherefore CBD fhall be 13; divide this by CB 5, there will 2 be found for BD. Whence AD will be found out by the 47. 1. PRO P. XIII. Fig. 57. In acute-angled triangles as ABC, the fquare made of the fide AB, fubtending the acute angle ACB, is less than the fquares angle made f #6. 13, a 47. 1. b 4. 2. € 47. I. a 47. I. b 7. 2. € 47. I. a 45. I. b 10. I. * 46. 1. 3. ax. e 47. 1. and 3. ax. made of the fides AC, CB, comprehending the acute angle I fay that ACq+BCq=ABq+2BCD. For these are (a) ADq+DCq+BCq. Coroll. Hence, the fides of an acute-angled triangle ABC being known, you may find out the fegment DC, intercepted betwixt the perpendicular AD, and the acute angle ACB, as also the perpendicular it felf AD. Let AB be 13, AC 15, BC 14. Take ABq (169) from ACq+BCq, that is, from 225-196421. Then remains 252 for 2BCD, wherefore BCD will be 126, divide this by BC 14, then will 9 be found out for DC. From whence it follows, AD√ 225—81—12. PROP. XIV. Plate I. Fig. 58. To find a fquare MC equal to a right-lined figure given A. For let GH be drawn. DCF (d) =GFq-GCq (e) was to be done. Then is A (c) =DB (c) = =HCq (c) MC. Which The End of the fecond Boo K. The |