22. Given two angles of a triangle, and the side adjacent to them, to construct the triangle. 23. Given the base of a triangle, the sum or difference of the sides, and one of the angles at the base, to construct the triangle. 24. Given two angles of a triangle, and the perimeter, to construct the triangle. 25. Given one leg of a right-angled triangle, and the sum or difference of the hypothenuse and the other leg, to construct the triangle. 26. Through a given point, to draw a straight line that shall make equal angles with two straight lines given in position. Prop. 27 to 34. 27. Prove either of the parts of I. 28 indirectly, and from it when proved, derive the other part, and also I. 27. 28. If two straight lines contain an angle, two other straight lines parallel to them will contain an equal angle. 29. Three straight lines parallel respectively to the three sides of a triangle form an equiangular triangle. 30. If a straight line joining two parallel straight lines be bisected, any other line drawn through the point of bisection, and terminated by the parallels, is bisected in that point. 31. From a given rhombus to cut off two isosceles triangles, such that the remaining hexagon shall be equilateral. 32. Within or without a given triangle, to draw a straight line parallel to the base, such that it may be equal to the parts of the sides or their continuations, between it and the base. 33. To trisect a given right angle. 34. If either the opposite sides, or the opposite angles of a quadrilateral be equal, it is a parallelogram. 35. If the diagonals of a quadrilateral bisect each other, it is a parallelogram, and if it be a parallelogram, its diagonals bisect each other. 36. To trisect a given straight line; and also, to cut it into any number of equal parts. 37. Given the two sides of a triangle, and the straight line drawn from the vertex to the point of bisection of the base, to construct the triangle. 38. Through a given point to draw a straight line, so that the parts of it intercepted between that point, and perpendiculars upon Prop. 35 to 48. 39. The straight line drawn from the vertex of a triangle to the point of bisection of the base, bisects any line parallel D to the base, and terminated by the sides, or the sides produced. And conversely, any line so bisected is parallel to the base. 40. To describe a triangle equal to a given quadrilateral figure. 41. If two triangles have two sides of the one equal respectively to two sides of the other, and have likewise the contained angles supplementary, the triangles are equal in area. 42. To bisect a triangle by a straight line drawn through a given point in one of the sides. 43. The straight lines joining the points of bisection of the three sides of a triangle divide it into four equal triangles, and each of the straight lines is parallel to a side of the triangle, and equal to half of it. 44. The straight lines joining the points of bisection of the sides of a trapezium form a parallelogram equal to half of the trapezium. 45. To describe a triangle equal to a given rectilineal figure having one of its angles equal to a given angle, and one of its sides equal to a given straight line. 46. Within a given square to inscribe another, having its side equal to a given straight line. What are the limits of possibility ? 47. In the figure for I. 47, (1.) produce DB to meet FG, and LA to meet FG produced, and prove the prop. by means of I. 35 and 36. (2.) Prove that FC and AD cut each other at right angles. (3.) Join FD, EK, and GH, and prove that the triangles FBD, ECK, GAH, are each equal to ABC. 48. Describe a square equal to the sum of two or more given squares; and also one equal to the difference of two given squares. 49. The difference of the squares of the sides of a triangle is equal to the difference of the squares of the segments of the base made by a perpendicular from the opposite angle. 50. The three straight lines bisecting the three angles of a triangle, all meet in the same point. 51. Describe a rhombus equal to a given square, and having one of its diagonals double of the other. 52. Divide a given parallelogram into five equal parts, by straight lines drawn through one of the angles. 53. The three straight lines drawn from the angular points of a triangle to the points of bisection of the opposite sides, pass through the same point, and trisect each other in that point. 54. Given the straight line drawn from the vertex of an equilateral triangle to one of the points of trisection of the base, to construct the triangle. BOOK II. DEFINITIONS. A D 1. A rectangle, or right-angled parallelogram, is said to 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 diagonal, 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. The rectangle under, or contained by two lines, as AB and BC, is concisely expressed thus, AB BC; and the square of a line AB may be written AB?. H K B G 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, anil the several parts of the divided line. (References-Prop. I. 3, 11, 31, 34.) Hypothesis.—Let A and BC be two straight lines; and let EC be divided into any parts in the points D, E. Sequence. The rectangle contained by the straight lines 4 and BC, shall be equal to the rectangle contained by A and BD, together with that contained by A and DE, and that contained by A and EC. B (I. 3.) G к A Construction.—1. From the point B draw BF at right angles to BC. (I. 11.) 2. And make BG equal to A. 3. Through G draw GH parallel to BC. (I. 31.) 4. And through the points D, E, C, draw DK, EL, CH, parallel to BG. (I. 31.) Demonstration.-1. Then the rectangle BH is equal to the rectangles BK, DL, EH. 2. But BH is contained by A and BC, for it is contained by GB and BC, and GB is equal to A. (const.). 3. And BK is contained by A and BD, for it is contained by GB and BD, and GB is equal to A. 4. And DL is contained by A and DE, because DK is equal to BG, which is equal to A. (I. 34.) 5. And in like manner EH is contained by A and EC. 6. 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. Conclusion. 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. (References—Prop. I. 31, 46.) Hypothesis.—Let the straight line AB be divided into any two parts in the point C. Sequence.—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.—1. Upon AB describe the square ADEB. (I. 26.) Through C draw CF parallel to AD or BE. (I. 31.) Demonstration.-1. Then AE is equal to the rectangles AF and CE. 2. But AE is the square on AB. A с D 3. Therefore the square on AB is equal to the rectangles AF and CE. 4. 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. 5. And CE is contained by AB and BC, for BE is equal to AB. 6. Therefore the rectangle AB AC, together with the rectangle AB BC, is equal to the square on AB. Conclusion.—Therefore, if a straight line, &c. Q. E. D. PROPOSITION 3.—THEOREM. 13 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. (References—Prop. I. 31, 46.) Hypothesis.—Let the straight line AB be divided into any two parts in the point C. Sequence.—The rectangle AB:BC shall be equal to the square on BC, together with the rectangle AC-CB. Construction.-1. Upon BC describe the square CDEB. (I. 46.) 2. Produce ED to F; and through A draw AF parallel to CD or BE. (I. 31.) Demonstration.-1. Then therectangle AE is equal to the rectangles AD and CE. 2. 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. 3. And AD is coutained by AC and CB, for CD is equal to CB. 4. And CE is the square on BC. 5. Therefore the rectangle AB BC is equal to the square on BC, together with the rectangle AC.CB. Conclusion. Therefore, if a straight line, &c. Q. E. D). D PROPOSITION 4.-THEOREM. If a straight line be divided into any two parts, the square on the whole line is equal to the sum of the squares on the two parts, together with twice the rectangle contained by the parts. |