36. The area of any trapezium is half the rectangle contained by one of the diagonals of the trapezium, and the sum of the perpendiculars let fall upon it from the opposite angles. 37. If the middle points of the sides of a triangle be joined, the lines form a triangle whose area is one-fourth that of the given triangle. 38. If the sides of a triangle be such that they are respectively the sum of two given lines, the difference of the same two lines, and twice the side of a square equal to the rectangle contained by these lines, the triangle shall be right-angled, having the right angle opposite to the first-named side. 39. If a point be taken within a triangle such that the lengths of the perpendiculars upon the sides are equal, show that the area of the rectangle contained by one of the perpendiculars and the perimeter of the triangle is double the area of the triangle. 40. In the last problem, if O be the given point, and OD, OE, OF the respective perpendiculars upon the sides BC, AC, and AB, show that the sum of the squares upon AD, OB, and DC exceeds the sum of the squares upon AF, BD, and CD by three times the square upon either of the perpendiculars. 41. Having given the lengths of the segments AF, BD, CE, in Problem 40, construct the triangle. 42. Draw a line, the square upon which shall be seven times the square upon a given line. 43. Draw a line, the square upon which shall be equal to the sum or difference of two given squares. 44. Reduce a given polygon to an equivalent triangle. 45. Divide a triangle into equal areas by drawing a line from a given point in a side. 46. Do the same with a given parallelogram. 47. If in the fig., Euc. I. 47, the square on the hypothenuse be on the other side, show how the other two squares may be made to cover exactly the square on the hypothenuse. 48. The area of a quadrilateral whose diagonals are at right angles is half the rectangle contained by the diagonals. 49. Bisect a given triangle by a straight line drawn from one of its angles. 50. Do the same with a given rectilineal figure ABCDEF. 51. If from the angle A of a triangle ABC a perpendicular be drawn meeting the base or base produced in D, show that the difference of the squares of AB and AC is equal to the difference of the squares of BD and DC. 52. If a straight line join the points of bisection of two sides of a triangle, the base is double the length of this line. 53. ABCD is a parallelogram, and E a point within it, and lines are drawn through E parallel to the sides of the parallelogram, show that E must lie on the diagonal AC when the figures BC and DE are equal. 51. If AD, BE, CF, are the perpendiculars from the angular points of the triangle ABC upon the opposite sides, show that the sum of the squares upon AE, CD, BF, is equal to the sum of the squares upon CE, BD, AF. 55. The diagonals of a parallelogram bisect one another. 56. Write out at full length a definition of parallelism, and then prove that the alternate angles are equal when a straight line meets two parallel straight lines. 57. ABCDE are the angular points of a regular pentagon, taken in order. Join AC and BD meeting in H, and show that AEDH is an equilateral parallelogram. 58. Having given the middle points of the sides of a triangle, show how to construct the triangle. 59. Show that the diagonal of a parallelogram diminishes while the angle from which it is drawn increases. What is the limit to which the diagonal approaches as the angle approaches respectively zero and two right angles? 60. A, B, C, are three angles taken in order of a regular hexagon, show that the square on AC is three times the square upon a side of the hexagon. EUCLID'S ELEMENTS, BOOK II. THE student, as he works through this book of Euclid's Elements, will not fail to perceive how easily many geometrical truths may be proved, and how concisely they may be expressed by means of Algebra. We have not space here to at all discuss the relation between Geometry and Algebra, but we will express a few of the propositions of this book in Algebraical symbols. 1. A rectangle, or right-angled parallelogram, is said tɔ be contained by any two of the straight lines which contain one of the right angles. 2. In any parallelogram, the figure which is composed of either of the parallelograms about a diameter, together with the two complements, is called a gnomon. Thus the parallelogram HG, together with the complements AF, FC, is a gnomon, which is briefly expressed by the letters AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon. H A F B G The rectangle under, or contained by two lines, as AB and BC, is concisely expressed thus:-AB, BC. Proposition 1.-Theorem. If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E; The rectangle contained by the straight lines A and BC A·BC= shall be equal to the rectangle contained by A and BD, A BD + together with that contained by A and DE, and that contained by A and EC. CONSTRUCTION. From the point B draw BF at right angles to BC (I. 11), And make BG equal to A (I. 3). Through G draw GH parallel to BC (I. 31), And through the points D,E,C,draw DK, EL, CH parallel to BG (I. 31). PROOF. Then the rectangle BH is equal to the rectangles BK, DL, EH. G B D F C A DE+ K Τα H F Since A + DL + But BH is contained by A and BC, for it is contained EH. by GB and BC, and GB is equal to A (Const.); And BK is contained by A and BD, for it is contained by GB and BD, and GB is equal to A; And DL is contained by A and DE, because DK is equal to BG, which is equal to A (I. 34) ; And in like manner EH is contained by A and EC; Therefore the rectangle contained by A and BC is equal to the several rectangles contained by A and BD, by A and DE, and by A and EC. Therefore, if there be two straight lines, &c. Q. E. D. Proposition 2.-Theorem. If a straight line be divided into any two parts, the rectangles contained by the whole line and each of its parts are together equal to the square on the whole line. Let the straight line AB be divided into any two parts in the point C; AB BC +AB AC =AB. For AB2 is the sum of its The rectangle contained by AB and BC, together with the rectangle contained by AB and AC, shall be equal to the square on AB. CONSTRUCTION.-Upon AB describe the Through, C draw CF parallel to AD or PROOF.-Then AE is equal to the rect- But AE is the square on AB; D C Therefore the square on AB is equal to the rectangles AF parts AF and CE. CE. And..= And AF is the rectangle contained by BA and AC, for it is contained by DA and AC, of which DA is equal to BA; And CE is contained by AB and BC, for BE is equal to AB. Therefore the rectangle AB, AC, together with the rect AB AC+ angle AB, BC, is equal to the square on AB. Therefore, if a straight line, &c. Q. E. D. AB BC. AB BC= For Proposition 3.-Theorem. If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts is equal to the square on that part, together with the rectangle contained by the two parts. Let the straight line AB be divided into any two parts in the point C; The rectangle AB BC shall be equal to the square on BC, together with the rectangle AC CB. CONSTRUCTION.-Upon BC describe the square CDEB (I. 46). Produce ED to F; and through A draw AF parallel to CD or BE (I. 31). PROOF. Then the rectangle AE is equal to the rectangles AD and CE. But AE is the rectangle contained by AB and BC, for it is contained by AB and BE, of which BE is equal to BC; D B And AD is contained by AC and CB, for CD is equal to CB, |