PRO P. XXXV. Plate I. Fig. 37. . Parallelograms, BCDA, BCFE, which fand upon the fame base BC, and between the fame parallels AF, BC, are equal one to the other. For AD (a) BC (a) EF, add DE common to both; (b) then is A E D F. But alfo AB («t :) DC, and the angle A (c) =CDF. (d) Therefore is the triangle ABE-DCF. Take away Dg E common to both triangles; (e) then is the Trapezium AB g D = CF; add B g C, common to both; (f) then is the parallelogram ABCD EBCF. Which was to be demonfrated. E The demonftration of any other cafes, is not uniike, but much more plain and easy. Schol. Fig. 7. If the fide AB, of a right-angled parallelogram ABCD, be conceived to be carried along perpendicularly thro' the whole line AC, or AC thro' the whole line AB, the area or content of the rectangle ABCD fhall be produc'd by that motion. Hence a rectangle is faid to be made by the drawing or multiplication of two contiguous fides. For example; let AB be supposed four foot, and AC three; draw three into four, there will be produced twelve fquare feet for the area of the rectangle. This being fuppofed, the dimenfion of any parallelqgram (EBCF) is found out by this theorem. For the area thereof is produced from the altitude BA drawn into the bafe BC. For the area of the rectangle AC parallelogram ERCF, is made by the drawing of BA into BC, therefore, &c. PROP. XXXVI. Fig 37. Parallelograms BCDA, GHF E, standing upon equal bafes BC, GH, and betwixt the fame parallels AF, BH, are equal one to the other. Draw BE and CF. Because BC (a) = GH(b) = B 3 Whence PRQ P. a 31. 1. a 34. 1. b 36. 1. an! 7. ax. € 34. I. a 37. I. b hyp• a 38. 1. b hyp• e J. ax. PROP. XXXVII. Plate I. Fig. 38. Triangles, BCA, BCD, ftanding upon the fame bafe BC, and between the fame parallels BC, EF, are equal the one to the other. (a) Draw BE parallel to CA, (a) and C F parallel to B D. Then is the triangle BCA (6) half Pgr. BCAE (c) half BDFC (6) ≈ BCD. Which was to be demonftrated. 1 PROP. XXXVIII. Fig. 38. Triangles, BCA, DFC, fet upon equal bases BC, DF, and between the fame parallels EF, BC, are equal the one to the other. Draw EB parallel to AC, and DB parallel to FC. Schol. If the bafe BC be greater than DF, then is the triangle BACDFC, and fo on the contrary. PRO P. XXXIX. Fig. 39. Equal triangles BCA, BCD, standing on the fame base BC, and on the fame fide are alfo between the fame paral·lels AD, BC. If you deny it, let another line AF be parallel to BC; and let CF be drawn. Then is the triangle CBF (a)= CBA (b) = CBD. (c) Which is abfurd. PROP. XL. Fig. 40. Equal triangles BCA, CFD, standing upon equal bafes BC, CF, and on the fame fide, are betwixt the fame parallels. If you deny it, let another line AH be parallel to BF, and let FH be drawn. Then is the triangle CFH (a) BCA (b)=CFD. (c) Which is abfurd. PRO P. XLI. Fig. 38. If a Pgr. AEBC have the fame bafe BC with the triangle BCE, and be between the fame parallels DE, BC, then is the Pgr. AEBC double to the triangle BCE. Let Let the line AC be drawn. Then is the triangle BCA (a) —BCE; therefore is the Pgr. ABCD (6) =2BCA (e) zBCE. Which was to be demonftrated. Schol. From hence may the area of any triangle BCE be found, for whereas the area of the Pgr. ABCD is produced by the altitude drawn into the base, therefore shall the area of a triangle be produced by half the altitude drawn into the bafe, or half the bafe drawn into the altitude; thus, if the base BC be 8, and the altitude 7, then is the area of the triangle BCE 28. PROP. XLII. Plate I. Fig. 41. To make a Pgr. ECGF equal to a triangle given ABC in an angle equal to a right-lined angle given. Through A (a) draw AG parallel to BC, (b) make the angle BCG the angle given; (c) bifect the base BC in E, and draw EF parallel to CG; then is the problem refolved. For (AE being drawn) the angle ECG is equal to the given angle by conftruction, and the triangle BAC (d) =2 AEC (c)= Pgr. ECGF. Which was to be done. PROP. XLIII. Fig. 9. In every Pgr. AGEL, the complements LD, GD, of those Pgrs. CK, KF, which fand about the diameter, are equal one to the other. For the triangle AEL (a)=AEG, and the triangle ADC (a) =ADB; and the triangle DEK (a) -DEF (6) Therefore the Pgr. LDDG. Which was to be demonftrated. PRO P. XLIV. Fig. 42. To a given right-line A, to apply a parallelogram FL, equal to a given triangle LBC, in a given angle C. (a) Make a Pgr. FD equal to the triangle LBC, fo that the angle GFE may be equal to C. Produce GF till FH be equal to the line given A. Through H (b) draw IL parallel to EF, which let DE produced meet in I, let DG produced meet with a right-line drawn from I through F in the point K; thro' K (6) draw KL parallel to GH, which let EF continued meet at M, and IH at L. Then fhall FL be the Pgr. required. B 4 For For the Pgr. FL (c) =FD= LBC; (d) and the angle PROP. XLV. Plate I. Fig. 43, and 44. Upon a right-line given FG, and in a given angle A, to make a Pgr. F L, equal to a right-lined figure given ABCD. Refolve the right-lined figure given into two triangles BAD, BCD, then (a) make a Pgr. FH-BAD, fo that the angle F may be equal to A. FI being produced, (a) make on HI the Pgr. ILBCD. Then is the Pgr. FL (6)=FH+-IL (c)=ABCD. Which was to be done. Schol. Hence is eafily found the excefs, whereby any rightlined figure exceeds a less right-lined figure. PROP. XLVI. Fig. 6. Upon a right-line given CD to defcribe a fquare DB. For, whereas the Angle C+D (c) two right-an- After the fame manner you may easily describe a rectangle contained under two right-lines given. PRO P. XLVII. Fig. 45. In right-angled triangles BAC, the fquare BE, which is made on the fide BC that fubtends the right-angle BAC, is equal to both the squares BG, CH, which are made in the fides AB, AC, containing the right-angle. Join AE, and AD; and draw AM parallel to CE. Becaufe the angle DBC (a)=FBA, add the augle ABC common to them both; then is the angle ABD— FBC. Morcover, AB (6)=FB, and BD (b) =BC; (c) therefore is the triangle ABD=FBC. But the Pgr. BM (d) 2 ABD, d 41. 1. e 6. ax. 2 ABD, and the Pgr. (d) BG= 2FBC (for GAC is one right-line, by Hypothefis, and 14. 1.) (e) therefore is the Pgr, BM-BG. By the fame way of argument is the Pgr. CM CH. Therefore is the whole BE (ƒ)BG+ £ 2. ax. CH. Which was to be demonftrated. Schol. This most excellent and useful theorem hath deferved the title of Pythagoras his theorem, because he was the inventor of it. By the help of which the addition and subtraction of fquares are performed; to which purpose serve the two following problems, PROBLEM I. Plate I. Fig. 46. To make one square equal to any number of squares given. Let three fquares be given, whereof the fides are AB, BC, CE. (a) Make the right-angle FBZ, having the fides infinite; and on them transfer AB and BC; join AC, then is ACq (b) ABq BCq. Then transfer AC from B to X, and CE the third fide given from B to E; join EX. (b) Then is EXq EBq (CEq) + BXq (ACq)(c) CEq + ABq + BCq. Which was to be demonstrated. PROBLEM II. Fig. 2. Two unequal right-lines being given AE, EG, to make a fquare equal to the difference of the trvo fquares of the given lines AE, EG. From the center E, at the distance of AE, describe a circle, and from the point G erect a perpendicular FG, meeting the circumference in F; and draw EF. (a) Then is EFq (EAq) = EGq + FGq, (b) ThereEAqEGq=GFq. Which was to be done. PROBLEM III. Fig. 45. Any two fides of a right-angled triangle ABC, being known, to find out the third. Let the fides AB, AC, encompaffing the right-angle, be, the one 6 foot, the other 8. Therefore, whereas ACq - ABq 64+ 36 100 BCq, thence is BC = √100 = 10, Now |