23. In the figure, Euc. Iv. 5. Let AF bisect the angle at A, and be produced to meet the circumference in G. Join GB, GC and find the center H of the circle inscribed in the triangle ABC. The lines GH, GB, GC are equal to one another.

24. Let ABC be any triangle inscribed in a circle, and let the perpendiculars AD, BE, CF intersect in G. Produce AD to meet the circumference in H, and join BH, CH. Then the triangle BHC may be shewn to be equal in all respects to the triangle BGČ, and the circle which circumscribes one of the triangles will also circumscribe the other. Similarly may be shewn by producing BE and CF, &c.

25. First. Prove that the perpendiculars Aa, Bb, Cc pass through the same point O, as Theo. 112, p. 171. Secondly. That the triangles Acb, Bea, Cab are equiangular to ABC. Euc. II. 21. Thirdly. That the angles of the triangle abc are bisected by the perpendiculars; and lastly, by means of Prob. 4, p. 71, that ab+be+ca is a minimum.

26. The equilateral triangle can be proved to be the least triangle which can be circumscribed about a circle.

27. Through C draw CH parallel to AB and join AH. Then HAC the difference of the angles at the base is equal to the angle HFC. Euc. III. 21, and HFC is bisected by FG.

28. Let F, G, (figure, Euc.iv. 5,) be the centers of the circumscribed and inscribed circles; join GF, GA, then the angle GAF which is equal to the difference of the angles GAD, FAD, may be shewn to be equal to half the difference of the angles ABC and ACB.

29. This Theorem may be stated more generally, as follows:

Let AB be the base of a triangle, AEB the locus of the vertex; D the bisection of the remaining arc ADB of the circumscribing circle; then the locus of the center of the inscribed circle is another circle whose center is D and radius DB. For join CD: then P the center of the inscribed circle is in CD. Join AP, PB; then these lines bisect the angles CAB, CBA, and DB, DP, DA may be proved to be equal to one another.

30. Let ABC be a triangle, having C a right angle, and upon AC, BC, let semicircles be described: bisect the hypotenuse in D, and let fall DE, DF perpendiculars on AC, BC respectively, and produce them to meet the circumferences of the semicircles in P, Q; then DP may be proved to be equal to DQ.

31. Let the angle BAC be a right angle, fig. Euc. iv. 4. Join AD. Then Euc. II. 17, note p. 155.

32. Suppose the triangle constructed, then it may be shewn that the difference between the hypotenuse and the sum of the two sides is equal to the diameter of the inscribed circle.

33. Let P, Q be the middle points of the arcs AB, AC, and let PQ be joined, cutting AB, AC in DE; then AD is equal to AE. Find the center O and join OP, QO.

34. With the given radius of the circumscribed circle, describe a circle. Draw BC cutting off the segment BAC containing an angle equal to the given vertical angle. Bisect BC in D, and draw the diameter EDF: join FB, and with center F and radius FB describe a circle: this will be the locus of the centers of the inscribed circle (see Theorem 33, supra.) On DE take DG equal to the given radius of the inscribed circle, and through G draw GH parallel to BC, and meeting the locus of the centers in H. H is the center of the inscribed circle.

35. This may readily be effected in almost a similar way to the preceding Problem.

36. With the given radius describe a circle, then by Euc. III. 34.

37. Let ABC be a triangle on the given base BC and having its vertical angle A equal to the given angle. Then since the angle at A is constant, A is a point in the arc of a segment of a circle described on BC. Let D be the center of the circle inscribed in the triangle ABC. Join DA, DB, DC: then the angles at B, C, A, are bisected. Euc. IV. 4. Also since the angles of each of the triangles ABC, DBC are equal to two right angles, it follows that the angle BDC is equal to the angle A and half the sum of the angles B and C. But the sum of the angles B and C can be found, because A is given. Hence the angle BDC is known, and therefore D is the locus of the vertex of a triangle described on the base BC and having its vertical angle at D double of the angle at A.

38. Suppose the parallelogram to be rectangular and inscribed in the given triangle and to be equal in area to half the triangle it may be shewn that the parallelogram is equal to half the altitude of the triangle, and that there is a restriction to the magnitude of the angle which two adjacent sides of the parallelogram make with one another.

39. Let ABC be the given triangle, and A'B'C' the other triangle, to the sides of which the inscribed triangle is required to be parallel. Through any point a in AB draw ab parallel to A'B' one side of the given triangle and through a, b draw ac, be respectively parallel to AC, BC. Join Ac and produce it to meet BC in D; through D draw DE, DF, parallel to ca, cb, respectively, and join EF. Then DEF is the triangle required.

40. This point will be found to be the intersection of the diagonals of the given parallelogram.

41. The difference of the two squares is obviously the sum of the four triangles at the corners of the exterior square.

42. (1) Let ABCD be the given square: join AC, at A in AC, make the angles CAE, CAF, each equal to one-third of a right angle, and join EF.

(2) Bisect AB any side in P, and draw PQ parallel to AD or BC, then at P make the angles as in the former case.

43. Each of the interior angles of a regular octagon may be shewn to be equal to three-fourths of two right angles, and the exterior angles made by producing the sides, are each equal to one fourth of two right angles, or one half of a right angle.

44. Let the diagonals of the rhombus be drawn; the center of the inscribed circle may be shewn to be the point of their intersection.

45. Let ABCD be the required square. Join O, O' the centers of the circles and draw the diagonal AEC cutting OO' in E. Then E is the middle point of OO' and the angle AEO is half a right angle.

46. Let the squares be inscribed in, and circumscribed about a circle, and let the diameters be drawn, the relation of the two squares is manifest. 47. Let one of the diagonals of the square be drawn, then the isosceles right-angled triangle which is half the square, may be proved to be greater than any other right-angled triangle upon the same hypotenuse.

48. Take half of the side of the square inscribed in the given circle, this will be equal to a side of the required octagon. At the extremities on the same side of this line make two angles each equal to three-fourths of two right angles, bisect these angles by two straight lines, the point at which they meet will be the center of the circle which circumscribes the octagon, and either of the bisecting lines is the radius of the circle. 49. First shew the possibility of a circle circumscribing such a figure, and then determine the center of the circle.

50. By constructing the figures and drawing lines from the center of

the circle to the angles of the octagon, the areas of the eight triangles may be easily shewn to be equal to eight times the rectangle contained by the radius of the circle, and half the side of the inscribed square.

51. Let AB, AC, AD, be the sides of a square, a regular hexagon and an octagon respectively inscribed in the circle whose center is O. Produce AC to E making AE equal to AB; from E draw EF touching the circle in F, and prove EF to be equal to AD.

52. Let the circle required touch the given circle in P, and the given line in Q. Let C be the center of the given circle and C' that of the required circle. Join CC, CQ, QP; and let QP produced meet the given circle in R, join RC and produce it to meet the given line in V. Then RCV is perpendicular to VQ. Hence the construction.

53. Let A, B be the centers of the given circles and CD the given straight line. On the side of CD opposite to that on which the circles are situated, draw a line EF parallel to CD at a distance equal to the radius of the smaller circle. From A the center of the larger circle describe a concentric circle GH with radius equal to the difference of the radii of the two circles. Then the center of the circle touching the circle GH, the line EF, and passing through the center of the smaller circle B, may be shewn to be the center of the circle which touches the circles whose centers are A, B, and the line CD.

54. Let AB, CD be the two lines given in position and E the center of the given circle. Draw two lines FG, HI parallel to AB, CD respectively and external to them. Describe a circle passing through E and touching FG, HI. Join the centers E, O, and with center O and radius equal to the difference of the radii of these circles describe a circle; this will be the circle required.

55. Let the circle ACF having the center G, be the required circle touching the given circle whose center is B, in the point A, and cutting the other given circle in the point C. Join BG, and through A draw a line perpendicular to BG; then this line is a common tangent to the circles whose centers are B, G. Join AC, GC. Hence the construction.

56. Let C be the given point in the given straight line AB, and D the center of the given circle. Through C draw a line CE perpendicular to AB; on the other side of AB, take CE equal to the radius of the given circle. Draw ED, and at D make the angle EDF equal to the angle DEC, and produce EC to meet DF. This gives the construction for one case, when the given line does not cut or touch the other circle.

57. This is a particular case of the general problem; To describe a circle passing through a given point and touching two straight lines given in position.

Let A be the given point between the two given lines which when produced meet in the point B. Bisect the angle at B by BD and through A draw AD perpendicular to BD and produce it to meet the two given lines in C, E. Take DF equal to DA, and on CB take CG such that the rectangle contained by CF, CA is equal to the square on CG. The circle described through the points F, A, G, will be the circle required. Deduce the particular case when the given lines are at right angles to one another, and the given point in the line which bisects the angle at B. If the lines are parallel, when is the solution possible?

58. Let A, B, be the centers of the given circles, which touch externally in E; and let C be the given point in that whose center is B. Make CD equal to AE and draw AD; make the angle DAG equal to the angle ADG: then G is the center of the circle required, and GC its radius.

59. If the three points be such as when joined by straight lines a triangle is formed; the points at which the inscribed circle touches the sides of the triangle, are the points at which the three circles touch one another. Euc. IV. 4. Different cases arise from the relative position of the three points.

60. Bisect the angle contained by the two lines at the point where the bisecting line meets the circumference, draw a tangent to the circle and produce the two straight lines to meet it. In this triangle inscribe a circle.

61. From the given angle draw a line through the center of the circle, and at the point where the line intersects the circumference, draw a tangent to the circle, meeting two sides of the triangle. The circle inscribed within this triangle will be the circle required.

62. Let the diagonal AD cut the arc in P, and let O be the center of the inscribed circle. Draw OQ perpendicular to AB. Draw PE a tangent at P meeting AB produced in E: then BE is equal to PD. Join PQ, PB. Then AB may be proved equal to QE. Hence AQ is equal to BE or DP.

63. Suppose the center of the required circle to be found, let fall two perpendiculars from this point upon the radii of the quadrant, and join the center of the circle with the center of the quadrant and produce the line to meet the arc of the quadrant. If three tangents be drawn at the three points thus determined in the two semicircles and the arc of the quadrant, they form a right-angled triangle which circumscribes the required circle.

64. Let AB be the base of the given segment, C its middle point. Let DCE be the required triangle having the sum of the base DE and perpendicular CF equal to the given line. Produce CF to H making FH equal to DE. Join HD and produce it, if necessary, to meet AB produced in K. Then CK is double of DF. Draw DL perpendicular to CK.

65. From the vertex of the isosceles triangle let fall a perpendicular on the base. Then, in each of the triangles so formed, inscribe a circle, Euc. IV. 4; next inscribe a circle so as to touch the two circles and the two equal sides of the triangle. This gives one solution: the problem is indeterminate.

66. If BD be shewn to subtend an arc of the larger circle equal to one-tenth of the whole circumference :-then BD is a side of the decagon in the larger circle. And if the triangle ABD can be shewn to be inscriptible in the smaller circle, BD will be the side of the inscribed pentagon.

67. It may be shewn that the angles ABF, BFD stand on two arcs, one of which is three times as large as the other.

68. It may be proved that the diagonals bisect the angles of the pentagon, and the five-sided figure formed by their intersection, may be shewn to be both equiangular and equilateral.

69. The figure ABCDE is an irregular pentagon inscribed in a circle; it may be shewn that the five angles at the circumference stand upon arcs whose sum is equal to the whole circumference of the circle; Euc. III. 20.

70. If a side CD (figure, Euc. IV. 11) of a regular pentagon be produced to K, the exterior angle ADK of the inscribed quadrilateral figure ABCD is equal to the angle ABC, one of the interior angles of the pentagon. From this a construction may be made for the method of folding the ribbon.

71. In the figure, Euc. Iv. 10, let DC be produced to meet the circumference in F, and join FB. Then FB is the side of a regular pentagon inscribed in the larger circle, D is the middle of the arc subtended by the adjacent side of the pentagon. Then the difference of FD and BĎ is equal to the radius AB. Next, it may be shewn, that FD is divided in the same manner in C as AB, and by Euc. II. 4, 11, the squares on FD and DB are three times the square on AB, and the rectangle of FD and DB is equal to the square on AB.

72. If one of the diagonals be drawn, this line with three sides of the pentagon forms a quadrilateral figure of which three consecutive sides are equal. The problem is reduced to the inscription of a quadrilateral in a square.

73. This may be deduced from Euc. Iv. 11.

74. The angle at A the center of the circle (fig. Euc. Iv. 10.) is onetenth of four right angles, the arc BD is therefore one-tenth of the circumference, and the chord BD is the side of a regular decagon inscribed in the larger circle. Produce DC to meet the circumference in F and join BF, then BF is the side of the inscribed pentagon, and AB is the side of the inscribed hexagon. Join FA. Then FCA may be proved to be an isosceles triangle and FB is a line drawn from the vertex meeting the base produced. If a perpendicular be drawn from F on BC, the difference of the squares on FB, FC may be shewn to be equal to the rectangle AB, BC, (Euc. 1. 47; 11. 5. Cor.); or the square on AC.

75. Divide the circle into three equal sectors, and draw tangents to the middle points of the arcs, the problem is then reduced to the inscription of a circle in a triangle.

76. Let the inscribed circles whose centers are A, B touch each other in G, and the circle whose center is C, in the points D, E; join A, D; A, E; at D, draw DF perpendicular to DA, and EF to EB, meeting in F. Let F, G be joined, and FG be proved to touch the two circles in G whose centers are A and B.

77. The problem is the same as to find how many equal circles may be placed round a circle of the same radius, touching this circle and each other. The number is six.

78. This is obvious from Euc. Iv. 7, the side of a square circumscribing a circle being equal to the diameter of the circle.

79. Each of the vertical angles of the triangles so formed, may be proved to be equal to the difference between the exterior and interior angle of the heptagon.

80. Every regular polygon can be divided into equal isosceles triangles by drawing lines from the center of the inscribed or circumscribed circle to the angular points of the figure, and the number of triangles will be equal to the number of sides of the polygon. If a perpendicular FG be let fall from F (figure, Euc. Iv. 14) the center on the base CD of FCD, one of these triangles, and if GF be produced to H till FH be equal to FG, and HC, HD be joined, an isosceles triangle is formed, such that the angle at H is half the angle at F. Bisect HC, HD in K, L, and join KL; then the triangle HKL may be placed round the vertex H, twice as many times as the triangle CFD round the vertex F.

81. The sum of the arcs on which stand the 1st, 3rd, 5th, &c. angles, is equal to the sum of the arcs on which stand the 2nd, 4th, 6th, &c. angles.

82. The proof of this property depends on the fact, that an isosceies triangle has a greater area than any scalene triangle of the same perimeter.