the planes and the other in the other, perpendicularly to the edge or common intersection. II. By the Translator. THE improvements referred to in the preceding note, so far as they have been adopted by the author, have been carefully preserved in the translation. Indeed, it has been found necessary, in a few instances, to use English words in a sense somewhat different from their ordinary acceptation. The word polygon is generally restricted to figures of more than four sides. It is used in this work with the latitude of the original word polygone to stand for rectilineal figures generally; and polyedron is adopted in a similar manner for solids. Quadrilateral is employed as a general name for four-sided figures. The word losenge is rendered by rhombus, and trapezé by trapezoid, the English words, as they are commonly used, corresponding to the French. The perpendicular let fall from the centre of a regular polygon upon one of its sides is called in the original apothéme. It occurs but a few times, and as there is no English word answering to it, it is rendered by a periphrasis, or simply by the word perpendicular. The portion of the surface of a sphere comprehended between the semicircumferences of two great circles is denoted in the original by fuseau; Dr. Hutton uses the word lune in the same sense; others have employed lunary surface; as lune properly stands for the surface comprehended between two unequal circular curves, the latter denomination was thought the least exceptionable, and is adopted in the translation. QUESTIONS IN GEOMETRY, INTENDED AS AN EXERCISE FOR THE LEARNER Q. I. From two given points A and B, on the same side of a line DE, Fig. 293 given in position, to draw two lines AP, PB, which shall meet in DE, and make equal angles with it. (34. 36.) Q. II. From two riven points A and B, to draw two equal straight lines Fig. 294. AE, BE, which shall meet in the same point of a line CD. (36 or 55.) Q. III. Through a given point P, to draw a line FE, which shall make Fig. 295. equal angles with two given straight lines BE, CF. (38.) Q. IV. If, from two points A and B, on the same side of a given line DE, Fig. 296. two straight lines AP, PB, be drawn, making equal angles with DE, AP and PB, will be together less than the sum of any other two lines AG, GB, drawn from A and B to any other point G in the line DE. (38.40.) Q. V. If the three sides of a triangle be bisected, the perpendiculars Fig. 297. drawn to the sides at the three points of bisection, will meet in the same point. Let the sides of the triangle ABC be bisected in the points D, E, F. Draw the perpendiculars EG, FG, meeting in G. The perpendicular at D also passes through G. (36. 48.) Q. VI. To divide a right angle into three equal parts.-Let ACB be Fig. 298. a right angle. Take CA of any magnitude, and erect upon it an equilateral triangle. (62.) Q. VII. Let ABC be an equilateral triangle. If the angles CBA and Fig. 299, CAB be bisected by the lines AD and BD, meeting in a point D, and DE, DF, be drawn parallel to the sides CA and CB respectively, the line AB will be divided into three equal parts at the points E and F. (76. 48.) Q. VIII. Any side of a triangle is greater than the difference of the Fig. 300. other two. Let AC be greater than AB, and cut off AD equal to AB. (45. 63.) Q. IX. If, from B, the vertex of the triangle ABC, BE be drawn perpen- Fig. 301 dicular to the base, and BD bisecting the angle ABC, the difference of the angles BAC and BCA is double the angle EBD. (57.) Q. X. If, from B one of the equal angles of an isosceles triangle, any Fig. 302. line BD be drawn to the opposite side, and from the same point B a line BE be drawn to the opposite side produced, so that DE shall be equal to DB, the angle ABD will be double the angle CBE. (63.45.) Geom. 29 Fig. 303 Fig. 304. Fig. 305. Fig. 306. Fig. 307. Fig 308. Fig. 309. Fig. 310. Fig. 311. Fig. 3.2. Q. XI. If, from the extremity C of the base BC of an isosceles triangle, a line CD equal to AC be drawn to meet the opposite side AB, produced if necessary, the angle DCE, formed by this line and the base produced, will be equal to three times the angle ABC. (45.63.) Q. XII. The sum of the sides of an isosceles triangle is less than the sum of the sides of any other triangle on the same base and between the same parallels. Let ACB be an isosceles triangle, and ADB any other triangle on the same base AB and between the same parallels AB and ED, AC and CB together will be less than AD and DB. `(76. Q. IV.) Q. XIII. If the three angles of the triangle ABC be bisected by the lines AD, BD, CD, and BD be produced to E, and if from D, DF be drawn perpendicular to AC, the angle ADE will be equal to CDF. (57.) Q. XIV. To bisect a parallelogram by a line drawn from a point in one of its sides. Let ABCD be a parallelogram, and P a given point in the side AB. Draw the diagonal BD, which bisects the parallelogram. Bisect BD in F, and through P and F draw PFE. PE bisects the parallelogram. Q. XV. If, in the sides of a square ABCD, four points E, F, G, H, be taken, one in each side, at equal distances from the four angles, the figure EFGH, contained by the straight lines which join these points, will be a square. Q. XVI. If the sides of the square described upon the hypothenuse of a right angled triangle be produced to meet the sides (produced if necessary) of the squares described upon the sides of the triangle, they will cut off triangles equiangular, and equal to the given triangle. Let DB, EC, the sides of the square described on BC, the hypothenuse of a right angled triangle ABC, be produced to meet the sides of the squares described upon BA, AC, in K and L; the triangles BFK, CIL, cut off by them, are equiangular, and equal to ABC. Q. XVII. To inscribe a square in a given right angled isosceles triangle. Divide the base AC of the right angled isosceles triangle into three equal parts. Q. XVIII. If the sides of an equilateral and equiangular pentagon be produced to meet, the angles formed by these lines are together equal to two right angles. Let the sides of the equilateral and equiangular pentagon ABCDE be produced to meet in the points F, G, H, I, K; the angles at these points are together equal to two right angles. (63.) Q. XIX. If the sides of an equilateral and equiangular hexagon, ABCDEF, be produced to meet in the points G, H, I, K, L, M, the angles at these points are together equal to four right angles. Q. XX. The perimeter of an isosceles triangle is greater than the perimeter of a rectangular parallelogram which is of the same altitude, and equal to the given triangle. Let ABC be an isosceles triangle whose base is BC. Draw AE perpendicular to BC, and therefore bisecting it, and draw AD, CD, parallel respectively to BC, AE; then DE is a rectangular parallelogram of the same altitude with, and equal to, the triangle ABC. The perimeter of ABC is greater than that of DE. Q. XXI. To determine a point in a line given in position, to which lines Fig. 319. drawn from two given points may have the greatest difference possible. Let A and B be the given points, CD the line given in position. Let fall the perpendicular BC, and produce it so that CE may be equal to CB; join AE, and produce it to meet CD in D. Join BD. D is the point required. (55. Q. VIII.) Q. XXII. Given one angle BCA, a side BC adjacent to it, and CD, the Fig. 300. difference of the other two sides, to construct the triangle. Q. XXIII. If, from a point P, without the circle ABC, two straight lines Fig. 314. PB, PD, be drawn to the concave part of the circumference, making equal angles with PO, the line joining P and the centre, the parts of the lines, AB, CĎ, which are intercepted within the circle, are equal. ` (38. 109.) Q. XXIV. Of all the straight lines which can be drawn from two given Fig. 315. points to meet on the convex circumference of a given circle, the sum of those two will be the least which make equal angles with the tangent at the point of meeting. Let A and B be two given points, CE a tangent to the circle at C, where the lines AC, BC, make equal angles with it, and AD, BD, be drawn from A and B to any other point D on the convex circumference; AC and CB are together less than AD and DB together. (Q. IV.) Q. XXV. If a straight line AB touch the interior of two circles, whose Fig. 316 common centre is O, in the point C, and terminate in the circumference of the exterior, it will be bisected at the point C. (110.) Q. XXVI. From two given points A and B, on the same side of a line Fig. 317 CD, to draw two straight lines AP, BP, which shall contain a given angle, and be terminated in that line. Q. XXVII. If any point which is not the centre be taken in the diameter Fig. 318. of a circle, of all the chords which can be drawn through that point, that is the least which is at right angles with the diameter. In AB, the diameter of the circle ADB, let any point C be taken which is not the centre, and let DE, FG, be any chords drawn through it, of which DE is perpendicular to AB; DE is less than FG. (109.) Q. XXVIII. If, from any point E, without a circle ACB, lines EA, EC, Fig. 319 be drawn touching it in the points A and C, and if EC meet the diameter drawn from A in the point D, then the angle AEC, contained by the tangents, will be double the angle CAB, contained by the line AC, which joins the points of contact, and AB the diameter drawn through one of them. Draw the diameter CF. (110.) Q. XXIX. If, from the extremities of the diameter of a circle, tangents Fig. 320. be drawn and produced to intersect a tangent to any point of the circumference, the straight lines joining the points of intersection and the centre of the circle form a right angle. Let A and B be the extremities of the diameter AB, let tangents AD, BE, be drawn meeting a tangent to any other point C of the circumference, in D and E; and let O be the centre, join DO, EO; the angle DOE is a right angle. Join CO. (110.) Q. XXX. In the diameter of a circle produced to determine a point from Fig. 321. which a tangent drawn to the circumference shall be equal to the diame ter. From A, the extremity of the diameter AB, draw AD at right angles, and equal to AB. Find the centre O, join OD, cutting the circle in C, and through C draw CE at right angles to OD meeting BA produced in E. |