PROP. VI. Plate I. Fig. 53. If a right-line A be divided into two equal parts, and another right-line E, added to the same 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 Square of AE taken as one line. 2 Q. a 4 & 3. I say that Aq. ((a) Q. A) +AE+Eq=Q. A+E. (a) For, Q. A+E=Aq+Eq AE. Which was to be demonftrated. Coroll. Hence it follows, that if 3 right-lines E, ETA, EA be in arithmetical proportion, then the rectangle he rectangle contained under the extreme terms E, ETA, together with the square 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 square made of one of the segments E, is equal to a double rectangle comprehended under the whole line Z, and the faid fegment E, together with the square 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 was to be demonstrated. Coroll. Hence it follows, that the square of the difference of any two lines Z, E, is equal to the square of both the lines less by a double rectangle comprehended under the faid lines. (c) For Zq+Eq-2ZE=Aq=Q. Z-E. PROP. 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 segments E four times, together with the square of the other fegment A, is equal to the square of the whole line Z, and the segment E, taken as one line Z+ E. I fay, that 4ZE+Aq=QZ + E. For 2 ZE (a)= Zq + Eq - Aq. Therefore 4 ZE + Aq=Zq+Eq +2ZE (6)=QZ+E. Which was to be demonftrated. PROP. a 4.2. b 3. 2. C 7. 2. and 3. ax. a 7. 2. and 3. ax. b 4. 2. 32 a 4. 2. bbyp. C 7. 2. d 2. ax. a 4.2. The Second Book of PROP. IX. Plate I. Fig. 54. If a right-line AB be divided into equal parts AC, CB and into unequal parts AD, DB, then are the squares of the unequal parts AD, DB, together, double to the square of the half line AC, and to the square of the difference CD. I say that ADq + DBq=2 ACq+2CDq. For ADq + DBq (a) = ACq + CDq + 2 ACD + DBq. But 2ACD (6) (2BCD) +DBq (c) = CBq (ACq) + CDq. (d) Therefore ADq+DBq= 2 ACq+2CDq. Which was to be demonstrated. This may be otherwise delivered and more easily demonstrated thus; the aggregate of the squares made of the fum and the difference of two right-lines A, E, is equal to the double of the squares made from those lines. For Q.A+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 demonstrated. PROP. X. Fig. 53. If a right-line A be divided into two 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 square of the added line E, double to the Square of half A, and the added line E, taken as one line. I say that Eq + Q. A + E, i. e. (a) Aq+2 Eq+2AE 2A2QA+E, For A+E, For 2 Q. A (6) Aq. And 2Q A+E (c) = Aq + 2 Eq + 2AE. Which was to be demonstrated, 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 segments BG, shall be equal to the Square, that is made of the other segment AG. Upon AB (a) defcribe the square AC (6) Bisect the fide AD in E, and draw the line EB; from the line EA produced take EF = EB. On AF make the square AH. Then is AH = ABX BG. For i : For HG being drawn out to I; the rectangle DH-+ EAq (c)=EFq (d) =EBq (e) =BAq+EAq: Therefore is DH (f) =BAq to the square AC. Take away AI common to both, there remains the square AH=GC, that is, AGq=ABX BG. Which was to be done. Schol. This proposition cannot be performed by numbers; * for there is no number that can be so divided, that the product of the whole into one part shall be equal to the square of the other part. PROP. XII. Plate I. Fig. 56. In obtufe-angled triangles as ABC, the square that is made of the fide AC, fubtending the obtuse angle ABC, is greater than the squares of the fides BC, AB, that contain the obtufe ang angle ABC, by a double rectangle contained under one of the fides BC, which are about the obtuse 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 obtuse angle ABC. I fay that ACq = CBq+ABq+2CB X BD. For these are all equal ACq. (CBq+2CBD+ (c) ABq. Scholium. Hence, the fides of any obtufe-angled triangle ABC being known, the segment 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 shall be 13; divide this by CB 5, there will 2 be found for BD. Whence AD will be found out by the 47. 1. PROP. XIII. Fig. 57. In acute-angled triangles as ABC, the square made of the fide AB, fubtending the acute angle ACB, is less than the squares angle made c 6. 2. d constr. e 47. 1. f 3. ax. *6.13. a 47.1. b 4. 2. C 47.1. • a 47. 1. b 7. 2. made of the fides AC, CB, comprehending the acute angle ACB, by a double rectangle contained under one of the fides BC, which are about the acute angle ACB, on which the perpendicular AD falls, and under the line DC, taken within the triangle from the perpendicular AD, to the acute. angle ACB. I fay that ACq+BCq=ABq+2BCD. For these are € 47. 1. equal. a 45. I. b 10. I. * 46. 1. c constr. d 5. 2. and 3. ax. e 47. 1. and 3. ax. ACq+BCq (a) ADq+DCq+BCq. (6) ADq+BDq+2BCD. (c) ABq+2BCD. Coroll. Hence, the fides of an acute-angled triangle ABC being known, you may find out the segment 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 225-196-421. 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 square 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 |