32. Having given the hypotenuse of a right-angled triangle, and the radius of the inscribed circle, to construct the triangle.

33. ABC is a triangle inscribed in a circle, the line joining the middle points of the arcs AB, AC, will cut off equal portions of the two contiguous sides measured from the angle 4.

IV.

34. Having given the vertical angle of a triangle, and the radii of the inscribed and circumscribed circles, to construct the triangle.

35. Given the base and vertical angle of a triangle, and also the radius of the inscribed circle, required to construct it.

36. Given the three angles of a triangle, and the radius of the inscribed circle, to construct the triangle.

37. If the base and vertical angle of a plane triangle be given, prove that the locus of the centers of the inscribed circle is a circle, and find its position and magnitude.

V.

38. In a given triangle inscribe a parallelogram which shall be equal to one-half the triangle. Is there any limit to the number of such parallelograms?

39. In a given triangle to inscribe a triangle, the sides of which shall be parallel to the sides of a given triangle.

40. If any number of parallelograms be inscribed in a given parallelogram, the diameters of all the figures shall cut one another in the same point.

41. A square is inscribed in another, the difference of the areas is twice the rectangle contained by the segments of the side which are made at the angular point of the inscribed square.

42. Inscribe an equilateral triangle in a square, (1) When the vertex of the triangle is in an angle of the square. (2) When the vertex of the triangle is in the point of bisection of a side of the square. 43. On a given straight line describe an equilateral and equiangular octagon.

VI.

44. Inscribe a circle in a rhombus.

45. Having given the distances of the centers of two equal circles which cut one another, inscribe a square in the space included between the two circumferences.

46. The square inscribed in a circle is equal to half the square described about the same circle.

47. The square is greater than any oblong inscribed in the same circle.

48. A circle having a square inscribed in it being given, to find a circle in which a regular octagon of a perimeter equal to that of the square, may be inscribed.

49. Describe a circle about a figure formed by constructing an equilateral triangle upon the base of an isosceles triangle, the vertical angle of which is four times the angle at the base.

50. A regular octagon inscribed in a circle is equal to the rectangle

contained by the sides of the squares inscribed in, and circumscribed about the circle.

51. If in any circle the side of an inscribed hexagon be produced till it becomes equal to the side of an inscribed square, a tangent drawn from the extremity, without the circle, shall be equal to the side of an inscribed octagon.

VII.

52. To describe a circle which shall touch a given circle in a given point, and also a given straight line.

53. Describe a circle touching a given straight line, and also two given circles.

54. Describe a circle which shall touch a given circle, and each of two given straight lines.

55. Two points are given, one in each of two given circles; describe a circle passing through both points and touching one of the circles. 56. Describe a circle touching a straight line in a given point, and also touching a given circle. When the line cuts the given circle, shew that your construction will enable you to obtain six circles touching the given circle and the given line, but not necessarily in the given point.

57. Describe a circle which shall touch two sides and pass through one angle of a given square.

58. If two circles touch each other externally, describe a circle which shall touch one of them in a given point, and also touch the other. In what case does this become impossible?

59. Describe three circles touching each other and having their centers at three given points. In how many different ways may this be done?

VIII.

60. Let two straight lines be drawn from any point within a circle to the circumference: describe a circle, which shall touch them both, and the arc between them.

61. In a given triangle having inscribed a circle, inscribe another circle in the space thus intercepted at one of the angles.

the

62. Let AB, AC, be the bounding radii of a quadrant; complete square ABDC and draw the diagonal AD; then the part of the diagonal without the quadrant will be equal to the radius of a circle inscribed in the quadrant.

63. If on one of the bounding radii of a quadrant, a semicircle be described, and on the other, another semicircle be described, so as to touch the former and the quadrantal arc; find the center of the circle inscribed in the figure bounded by the three curves.

64. In a given segment of a circle inscribe an isosceles triangle, such that its vertex may be in the middle of the chord, and the base and perpendicular together equal to a given line.

65. Inscribe three circles in an isosceles triangle touching each other, and each of them touching two of the three sides of the triangle.

IX.

66. In the fig. Prop. 10, Book IV, shew that the base BD is the

side of a regular decagon inscribed in the larger circle, and the side of a regular pentagon inscribed in the smaller circle.

67. In the fig. Prop. 10, Book IV, produce DC to meet the circle in F, and draw BF; then the angle ABF shall be equal to three times the angle BFD.

68. If the alternate angles of a regular pentagon be joined, the figure formed by the intersection of the joining lines will itself be a regular pentagon.

69. If ABCDE be any pentagon inscribed in a circle, and AC, BD, CE, DA, EB be joined, then are the angles ABE, BCA, CDB, DEC, EAD, together equal to two right angles.

70. A watch-ribbon is folded up into a flat knot of five edges, shew that the sides of the knot form an equilateral pentagon.

71. If from the extremities of the side of a regular pentagon inscribed in a circle, straight lines be drawn to the middle of the arc subtended by the adjacent side, their difference is equal to the radius ; the sum of their squares to three times the square of the radius; and the rectangle contained by them is equal to the square of the radius.

72. Inscribe a regular pentagon in a given square so that four angles of the pentagon may touch respectively the four sides of the

square.

73. Inscribe a regular decagon in a given circle.

74. The square described upon the side of a regular pentagon in a circle, is equal to the square on the side of a regular hexagon, together with the square upon the side of a regular decagon in the same circle.

X.

75. In a given circle inscribe three equal circles touching each other and the given circle.

76. Shew that if two circles be inscribed in a third to touch one another, the tangents of the points of contact will all meet in the same point.

77. If there be three concentric circles, whose radii are 1, 2, 3; determine how many circles may be described round the interior one, having their centers in the circumference of the circle, whose radius is 2, and touching the interior and exterior circles, and each other.

78. Shew that nine equal circles may be placed in contact, so that a square whose side is three times the diameter of one of them will circumscribe them.

XI.

79. Produce the sides of a given heptagon both ways, till they meet, forming seven triangles; required the sum of their vertical angles.

80. To convert a given regular polygon into another which shall have the same perimeter, but double the number of sides.

81. In any polygon of an even number of sides, inscribed in a circle, the sum of the 1st, 3rd, 5th, &c. angles is equal to the sum of the 2nd, 4th, 6th, &c.

82. Of all polygons having equal perimeters, and the same number of sides, the equilateral polygon has the greatest area.

HINTS, &c.

8. This is a particular case of Euc. 1. 22. The triangle however may be described by means of Euc. 1. 1. Let AB be the given base, produce AB both ways to meet the circles in D, E (fig. Euc. 1. I.); with center A, and radius AE, describe a circle, and with center B and radius BD, describe another circle cutting the former in G. Join GA, GB.

9. Apply Euc. 1. 6, 8.

10. This is proved by Euc. 1. 32, 13, 5.

11. Let fall also a perpendicular from the vertex on the base. 12. Apply Euc. 1. 4.

13. Let CAB be the triangle (fig. Euc. 1. 10.) CD the line bisecting the angle ACD and the base AB. Produce CD, and make DE equal to CD, and join AE. Then CB may be proved equal to AE, also AE to AC.

14. Let AB be the given line, and C, D the given points. From C draw CE perpendicular to AB, and produce it making EF equal to CE, join FD, and produce it to meet the given line in G, which will be the point required.

15. Make the construction as the enunciation directs, then by Euc. 1. 4, BH is proved equal to CK: and by Euc. 1. 13, 6, OB is shewn to be equal to OC.

16. This proposition requires for its proof the case of equal triangles omitted in Euclid:-namely, when two sides and one angle are given, but not the angle included by the given sides.

17. The angle BCD may be shewn to be equal to the sum of the angles ABC, AĎC.

18. The angles ADE, AED may be each proved to be equal to the complements of the angles at the base of the triangle.

19. The angles CAB, CBA, being equal, the angles CAD, CBE are equal, Euc. I. 13. Then, by Euc. 1. 4, CD is proved to be equal to CE. And by Euc. 1. 5, 32, the angle at the vertex is shewn to be four times either of the angles at the base.

20. Let AB, CD be two straight lines intersecting_each_other_in E, and let P be the given point, within the angle AED. Draw EF bisecting the_angle AED, and through P draw PGH parallel to EF, and cutting ED, EB in G, H. Then EG is equal to EH. And by bisecting the angle DEB and drawing through P a line parallel to this line, another solution is obtained. It will be found that the two lines are at right angles to each other.

21. Let the two given straight lines meet in A, and let P be the given point. Let PQR be the line required, meeting the lines AQ, AR in Q and R, so that PQ is equal to QR. Through P draw PS parallel to AR and join RS. Then APSR is a parallelogram and AS, PR the diagonals. Hence the construction.

22. Let the two straight lines AB, AC meet in A. In AB take any point D, and from AC cut off AE equal to AD, and join DE. On DE, or DE produced, take DF equal to the given line, and through F draw FG parallel to AB meeting AC in G, and through G draw GH parallel to DE meeting AB in H. Then GH is the line required.

23. The two given points may be both on the same side, or one point may be on each side of the line. If the point required in the line be supposed to be found, and lines be drawn joining this point and the given points, an isosceles triangle is formed, and if a perpendicular be drawn on the base from the point in the line: the construction is obvious.

24. The problem is simply this-to find a point in one side of a triangle from which the perpendiculars drawn to the other two sides shall be equal. If all the positions of these lines be considered, it will readily be seen in what case the problem is impossible.

25. If the isosceles triangle be obtuse-angled, by Euc. 1. 5, 32, the truth will be made evident. If the triangle be acute-angled, the enunciation of the proposition requires some modification.

26. Construct the figure and apply Euc. 1. 5, 32, 15.

=

If the isosceles triangle have its vertical angle less than two-thirds of a right-angle, the line ED produced, meets AB produced towards the base, and then 3. AEF 4 right angles +AFE. If the vertical angle be greater than two-thirds of a right angle, ED produced meets AB produced towards the vertex, then 3. AEF 2 right angles + AFE. 27. Let ABC be an isosceles triangle, and from any point D in the Dase BC, and the extremity B, let three lines DE, DF, BG be drawn to the sides and making equal angles with the base. Produce ED and make DH equal to DF and join BH.

28. In the isosceles triangle ABC, let the line DFE which meets the side AC in D. and AB produced in E, be bisected by the base in the point E. Then DC may be shewn to be equal to BE.

29. If two equal straight lines be drawn terminated by two lines which meet in a point, they will cut off triangles of equal area. Hence the two triangles have a common vertical angle and their areas and bases equal. By Euc. 1. 32 it is shewn that the angle contained by the bisecting lines is equal to the exterior angle at the base.

30. There is an omission in this question. After the words" making equal angles with the sides," add, "and be equal to each other respectively." (1), (3) Apply Euc. 1. 26, 4. (2) The equal lines which bisect the sides may be shewn to make equal angles with the sides.

31. At C make the angle BCD equal to the angle ACB, and produce AB to meet CD in D.

32. By bisecting the hypotenuse, and drawing a line from the vertex to the point of bisection, it may be shewn that this line forms with the shorter side and half the hypotenuse an isosceles triangle.

33. Let ABC be a triangle, having the right angle at A, and the angle at C greater than the angle at B, also let AD be perpendicular to the base, and AE be the line drawn to E the bisection of the base. Then AE may be proved equal to BE or EC independently of Euc. III. 31.

34. Produce EG, FG to meet the perpendiculars CE, BF, produced if necessary. The demonstration is obvious.

35. If the given triangle have both of the angles at the base, acute angles; the difference of the angles at the base is at once obvious from Euc. 1. 32. If one of the angles at the base be obtuse, does the property noid good?

36. Let ABC be a triangle having the angle ACB double of the angle ABC, and let the perpendicular AD be drawn to the base BC. Take DE equal to DC and join AE. Then AE may be proved to be equal to EB.

If ACB be an obtuse angle, then AC is equal to the sum of the segments of the base, made by the perpendicular from the vertex A.

37. Let the sides AB, AC of any triangle ABC be produced, the ex