PROB. XV. To describe a circle about a triangle ABC or (which is the same thing) thro' any three points, A, B, C, which are not situate in a right line. fig. 55. C fig. 55. H B By prob. 4. Bisect the line AC by the perpendicular DE, and also CB, by the perpendicular FG, the point of intersection H, of these perpendiculars, is the centre of the circle required, from which take the distance to any of the three points A, B, C, and desscribe the circle ABC, and it is done. For, by cor. to theo. 8. The lines DE and FG, must each pass thro' the centre, therefore, their point of intersection, H, must be the centre. SCHOLIUM. By this method the centre of a circle may be found by having only a segment of it given. PROB. XVI. To make an angle of any number of degrees, at the point A, of the line AB, suppose of 45 degrees. fig. 56. fig 56. From a scale of chords take 60 degrees, for 60 is equal to the radius (by cor. theo. 15.) and with that distance from A, as a centre, describe a circle from the line AB; take 45 degrees the quantity of the given angle, from the same scale of chords, and lay it on that circle from a to b, thro' A and b, draw the line AbC; and the angle A, will be an angle of 45 degrees, as required. If the given angle were more than 90; take its half (or divide it into any two parts less than 90) and lay them after each other on the arc which is described with the chord of 60 degrees, thro' the extremity of which, and the centre, let a line be drawn, and that will form the angle required, with the given line. PROB. XVII. fig. 57. To measure a given angle ABC. fig. 57. If the lines which include the angle, be not as long or a greater length, and between them so produced, with the chord of 60 from B, describe the arc e d, which distance e d, measured on the same line of chords, gives the quantity of the angle BAC 48 degrees, as required; this is plain from def. 19. PROB. XVIII. To make a triangle BCE equal to a given quadrilateral figure ABCD. fig. 58. Draw the diagonal AC, and parallel to it (by prob. 8.) DE, meeting AB produced in E, then draw CE, and ECB will be the triangle required. For the triangles ADC, AEC, being upon the same base AC, and under the same parallel ED (by cor. to theo. 13.) will be equal, therefore if ABC be added to each, then ABCD-BEC. PROB. XIX. To make a triangle DFH, equal to a given fivesided figure ABCDE. fig. 59. Draw DA and DB, and also EH and CF, parallel to them (by prob. 8.) meeting AB produced in H and F; then draw DH, DF, and the triangle HDF is the one required. For the triangle DEA=DHA, and DBC=DFB (by cor. to theo. 13.) therefore by adding these equations, DEA+DBC=DHA+DFB if to each of these ADB be added; then DEA+ADB+DBC=ABCDE =(DHA+ABD+DFB)=DHF. PROB. XX. To project the lines of chords, sines, tangents, and secants, to any radius. fig. 60. On the line AB, let a semicircle ADB be described: let CD be drawn perpendicular to the centre C, and the tangent BE perpendicular to the end of the diameter; let the quadrants AD, DB, be each divided into 9 equal parts, every one of which will be 10 degrees; if then from the centre C, lines be drawn thro' 10 20, 30, 40, &c. the divisions of the quadrant BD, and continued to BF, we shall there have the tangents of 10, 20, 30, 40, &c. and the secants C 10, C 20, C 30, &c. are transferred to the line CD, produced by desscribing the arcs, 10, 10: 20, 20: 30, 30, &c. If from 10, 20, 30, &c. the divisions of the quadrant BD, there be let fall perpendiculars, let these be transferred |
