(g) FGq + GDq equal. FGq+(6) GEq+Rectang. CED. (4) Therefore the rectangle AEB-CED. 4. Cafe. If neither of the right-lines AB, CD pafs thro' the center, then through the point of interfection E, draw the diameter GH. By that which hath been already demonftrated it appears, that the rectangle AEB=GEH =CED. Which was to be demonftrated. More easily, and generally, thus; join AC and BD, then because the angles (a) CEA, DEB, and (b) alfo C, B (upon the fame arch AD) are equal, thence are the triangles CEA, BED, (c) equiangular. (d) Wherefore CE : EA EB: ED, and (e) confequently CEX ED-AEX EB. Which was to be demonftrated. The citations out of the 6 Book, both here and in the following prop, have no dependance on the fame; so that it was free to use them. PROP. XXXVI. Plate II. Fig. 42. If any point be taken without a circle EBC, and from that point two right-lines DA, DB, fall upon the circle, whereof one DA cuts the circle, the other DB touches it, the rectangle comprehended under the whole line DA that cuts the circle, and DC, that part which is taken from the point given D to the convex of the periphery, fhall be equal to the fquare made of the tangent line. 1. Cafe. If the fecant AD paffes thro' the center, then join EB, this (a) will make a right-angle with the line DB, wherefore DBq+ (d) EBq (ECq,) (b) =EDq (c) = ADX DCECq. Therefore AD XDC=DBq. Which was to be demonstrated. 2. Cafe. But if AD paffes not thro' the center, then Fig. 43. draw EC, EB, ED, and EF perpendicular to AD, (a) wherefore AC is bifected in F. = Becaufe BDq+EBq (6) ≈DEq (6) EFq+FDq (c) EFq-t-ADC+FCq (d)=ADC+CEq (EBq.) (e) Therefore is BDq ADC. Which was to be demonftrated. More eafily, and generally thus; draw AB and BC. Then, because the angles A, and DBC (a) are equal, and the angle D common to both; thence are the triangles BDC, ADB (b) equiangular. (c) Wherefore AD: DB DB: CD; and (d) confequently ADXDC=DBq: Which was to be demonftrated. a 3. 3. b 47. I. e 3. ax. a 32. 3. a 36. 3. a 36. 3. b 36. 3. C 2. cor. d 8. 3. e 2. cor. f hyp. g 8.3. a 17. 3. b hyp. c 36. 3. dx. ax. & sch. 4. 1. e 8. I. f 12. ax. Coroll. Plate II. Fig. 45: 1. Hence, If from any point A, taken without a circle, there be several lines AB, AC drawn which cut the circle; the rectangle comprehended under the whole lines AB, AC, and the outward parts AL, AF, are equal between themselves. g cor. 16. 3. h 8. 1. For if the tangent AD be drawn, then is CAF-ADq (a) =BAE. 2. It appears alfo from hence, that if two lines AB, AC, (Fig. 46.) drawn from the fame point do touch a circle, thofe two lines are equal one to the other. For if AE be drawn cutting the circle, then is ABq (a) = EAF (b) =ACq, 3. It is alfo evident, that from a point A, taken without a circle, there can be drawn but two lines AB, AC, that fhall touch the circle. For if a third line AD be faid to touch the circle, thence is AD ()=AB (c)=AC. (d) Which is abfurd. 4. And, on the contrary, it is plain, that if two equal right-lines A B, A C, fall from any point A, upon the convex periphery of a circle, and that if one of these equal lines A B touch the circle, then the other A C touches the circle alfo. For if poffible, let not A C, but another line A D, touch the circle; therefore is AD (e) =AC (ƒ) =AB. (g) Which is abfurd. PROP. XXXVII. Fig. 47. If without a circle EBF any point D be taken, and from that point two right-lines DA, DB fall on the circle, whereof one line DA cuts the circle, the other DB falls upon it; and if alfo the rectangle comprehended under the whole line that cuts the circle, and under that part of it DC which is taken betwixt the point D and the convex periphery, be equal to that fquare which is made of the line DB falling on the circle, I fay that that line DB fo falling fhall touch the circle given. From the point D (a) let a tangent DF be drawn, and from the center E draw ED, EB, EF. Now because DBq (6) ADC (c)=DFq, therefore is DB (d)=DF: But EB-EF, and the fide ED common; (e) therefore the angle EBD EFD; but EFD is a right-angle, and (ƒ) therefore EBD is right also; and (g) therefore DB touches the circle. Which was to be dem. Coroll. From hence it follows that the (b) angle EDB EDF, The FOURTH BOOK O F EUCLIDE's I. ELEMENT S "Α Definitions. Right-lined figure is faid to be infcribed in a right-lined figure, when every one of the angles of the infcribed figure touch every one of the fides of the figure wherein it is inscribed. So the triangle DEF is infcribed in the triangle ABC. Plate II. Fig. 48. II. In like manner a figure is faid to be described about a figure, when every one of the fides of the figure circumfcribed, touch every one of the angles of the figure about which it is circumfcribed. So the triangle ABC is defcribed about the triangle DEF. III. A right-lined figure is faid to be infcribed in a circle, when all the angles of that figure which is inscribed touch the circumference of the circle. As Fig. 50. IV. A right-lined figure is faid to be described about a circle, when all the fides of the figure which is circumfcribed touch the periphery of the circle. As Fig. 49. V. After the like manner a circle is faid to be infcribed in a right-lined figure, when the periphery of the circle touches all the fides of the figure, in which it is infcribed. Fig. 49. VI. A circle is faid to be described about a figure when the periphery of the circle touches all the angles of the figure, which it circumfcribes. VII. A right-line is faid to be fitted or applied in a circle when the extremes thereof fall upon the circumference; as the right-line AB. Fig. 6. PROP. I. Fig. 51. In a circle given ABC to apply a right-line AB equal to a D 2 rights a 3. post. & 3. I. b 15. def. 1. c conftr. a 17. 3. b23. I. c 32. 3. d constr. e 32. I. a 23. 1. b 17. 3. c 13. ax dII. e sch, 32. 1. h 3. ax. 2 9. I biz. b 12. right-line given D, which doth not exceed AC, the diameter of the circle. From the center A at the the distance AED (a) describe a circle meeting with the circle given in B, draw AB. Then is AB (6) AE (c)—D. Which was to be done. PROP. II. Plate II. Fig. 49, 50. In a circle given ABC to defcribe a triangle ABC, equiangular to a triangle given DEF. Let the right-line GH (a) touch the circle given in A; (6) make the angle HACE, (6) and the angle GAB=F, then join BC; and the thing is done. For the angle B (c)=HAC (d)=E, and the angle C (c) =GAB (d) =F; (e) whence alfo the angle BACD. Therefore the triangle B A C inscribed in the circle is equiangular to DEF. Which was to be done. PROP. III. Fig. 52, 53. About a circle given IABC to describe a triangle LNM, equiangular to a triangle given DEF. Produce the fide EF on both fides; at the center I (a) make an angle AIB-DEG, and an angle BIC=DFH. Then through the points A, B, C, let three right-lines LN, LM, NM, (b) be drawn, touching the circle, and the thing is done. For it's evident that the right-lines LN, LM, MN, will meet and make a triangle, (c) because the angles LAI, LBI are right; fo that if the (d) right-line AB was drawn, it would make the angles LAB, LBA, less than two right-angles. Since therefore the angle AIB-+L (e)=2 right-angles (f)=DEG+DEF, and AIB (g)=DEG; (b) therefore is the angle L-DEF. By the fame way of reasoning the angle MD FE, (k) Therefore alfo the angle N= D. And therefore the triangle LNM, described about the circle, is equiangular to EDF, the triangle given. Which was to be done. PRO P. IV. Fig. 54. In a triangle given ABC, to defcribe a circle EFG. (a) Bisect the angles B and C with the right-lines BD, CD, meeting in the point D, (b) and draw the perpendiculars DE, DF, DG. A circle defcribed from the center center D through E, will pafs through G and F, and touch the three fides of the triangle. For the angle DBE (c) =DBF; and the angle DEB (d) =DFB; and the fide DB common, (e) therefore DE= DF. For the fame reafon DG-DF, The circle therefore defcribed from the center D paffes through the three points E, F, G; and whereas the angles at E, F, G, are right, therefore it touches all the fides of the triangle. Which was to be done. Schol. c conftr. diz ax, e 26. 1. Hence, The fides of a triangles being known, their feg. Pet. Herig. ments which are made by the touching of the circle inscribed, fhall be found, Thus ; Let AB be 12, AC 18, BC 16, then is AB+BC=28. Out of which fubduct 18 AC AE+FC, there remains 10 BE+BF. Therefore BE, or BF = 5; and confequently FC, or C G11. Wherefore GA, or AE, = = 7. PROP. V. Plate II. Fig. 55, 56. About a triangle given ABC, to describe a circle FABC. (a) Bisect any two fides BA, CA, with perpendiculars DF, EF, meeting in the point F. I fay this fhall be the center of the circle, Now For, let the right-lines FA, FB, FC be drawn. because AD (b)=DB, and the fide DF common, and the angles FDA (c) =FDB, therefore is FB (a) = FA. After the fame manner is FC FA. Therefore a circle described from the center F fhall pass through the angles of the triangle given (viz.) B, A, C. Which was to be done. Coroll. * Hence, if a triangle be acute-angled, the center « fhall fall within the triangle; if right-angled, in the fide oppofite to the right-angle, and if obtufe-angled, without the triangle. Schol. By the fame method may a circle be defcribed, that fhall pass through three points given, not being in the fame ftrait-line, D 3 PROP. |