7 22 ficient precision by multiplying the diameter by 22, and dividing the product by 7, which is the approximation discovered by Archimedes. The fraction is equal to 31428, and consequently the circumference, as determined by this last method, differs from the truth by rather more than a thousandth part of the diameter, which in most practical cases is too inconsiderable to deserve notice. BOOK VIII. · PROPOSITION I. PROBLEM. To divide a given straight line into any proposed number of equal parts. Let it be proposed to divide the straight line AB into a certain number of equal parts. From one extremity A draw an indefinite straight line AC, making any angle with AB, and upon it repeat one more than the proposed number of equal distances; then, supposing the last to terminate in C, and the last but one in G, the line AG will be di vided into the same number of equal parts that AB is to be divided into; let the point E be at the distance of two of those parts from C, then if CBD be drawn, making BD=CB, and the points D,E be joined, the line AB will be divided in F; so that BF will be one of the required parts of AB. For draw GB; then since CG-GE and CB=BD, the sides CE, CD of the triangle CED are divided proportionally by the line GB; therefore (Prop. V. B. VI.) GB is parallel to ED or EF: therefore, in the triangle AGB, we have the proportion AG: EG :: AB: FB, but AG is a given multiple of EG; therefore (Prop. XI. B. V.) AB is the same multiple of FB. Otherwise as follows:-From one extremity A draw the indefinite straight line AC, making A. any angle with AB, and from the other extremity draw BD, making an equal angle with BA. Upon BD repeat the distance AC as many times, wanting one, E B A as there are to be divisions of AB: draw CD, which will cut off from AB one of the required parts AE. For, since the angles A, B are equal, the triangles EAC, EBD are similar; therefore AC: BD :: AE: EB; hence, whatever multiple BD is of AC, the same multiple is EB of AE, that is, AÊ is one of the proposed parts of AB. PROPOSITION II. PROBLEM. To find a mean proportional between two given straight lines. Let the given lines be A and B. AB PROPOSITION III. PROBLEM. To find a fourth proportional to three given straight lines. From any point A draw two straight Join D, F, and parallel to DF draw E B AD AE :: AF: AG. D A F Cor. The same construction serves for finding a third proportional to two given lines, as A and B; this being the same as a fourth proportional to the three lines A, B, B. PROPOSITION IV. PROBLEM. To divide a given straight line AB into parts proportional to given lines. E Draw AC equal to the longest of the proposed lines, make AG equal to the line next in length, AF equal to the next, and so on:-join BC, and draw GE, FD, &c. parallel to BC, and the line AB will be divided by them as required; for these pa- B rallels cut the sides AB, AC of the triangle ABC proportionally (Prop. VI. Čor. B. VI.). PROPOSITION V. D A PROBLEM. } A straight line being given, to divide it so that the rectangle of the two parts may be equivalent to a given rectangle; or to prolong it, so that the rectangle contained by the whole line and the part added may be equivalent to a given rectangle. Let AB be the proposed straight line. Then, from the extremities A, B draw the perpendiculars AD, BE, equal to the sides of the given rectangle, and both upon the same side of AB if it is to be divided, but one on each side if it is to be prolonged: draw DE, on which as a diameter describe a circle meeting AB, or its extension in the point C; AC and CB are the parts required. For draw DC and CE. Then, the angle DCE being contained in a semicircle, is a right angle (Prop. XIV. Cor. 3. B. III.); and, therefore, in E D both cases of the problem, the angles ACD, BCE are together equal to a right angle. But the angles ACD, CDA are likewise together equal to a right angle; and, consequently, the angles BCE, CDA are equal. Wherefore the right angled triangles CBE, CAD are similar; whence AC: AD :: BE: CB, and, therefore, AC CB AD.BE. Scholium. It is obvious that in the second case of this problem, since a portion of the circle lies on each side of the line AB, the circumference must always intersect its extension in two points, C and C. But, in the first case, the circle may either cut AB in two points C and C', or touch it in a single point, which will hence mark a limitation of the problem. When the circle does not reach AB, the problem fails, thus intimating that the proposed line cannot be divided as required. PROPOSITION VI. PROBLEM. To construct a square that shall be equivalent to a given polygon. Reduce the proposed polygon AB to an equivalent rectangle (Prop. XVI. Cor. B. IV.), of which let A, B be the sides; draw a straight line CDE, making CD -A and DE-B: describe a semicircle on CE, and draw DF perpendicular to CE, termi D E nating in the arc; DF will be the side of the square sought, for (Prop. XXIII. Cor. 2. B. VI.) DF2=CD.DE. PROPOSITION VII. PROBLEM. To construct a square that shall be to a given square, AC, as the line E is to the line F. Draw an indefinite straight line GH, upon which take GK E and KH F; describe on GH a semicircle, and draw the |