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 base 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. (1) When the two lines are drawn perpendicular to the sides; apply Euc. 1. 26, 4. (2) The equal lines which bisect the sides of the triangle may be shewn to make equal angles with the sides. (3) When the two lines make equal angles with the sides; apply Euc. 1. 26, 4. 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 hold 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 terior angles bisected by two lines which meet in D, and let AD be joined, then AD bisects the angle BAC. For draw DE perpendicular on BC, also DF, DG perpendiculars on AB, AC produced, if necessary. Then DF may be proved equal to DG, and the squares on DF, DA are equal to the squares on DG, GA, of which the square on FD is equal to the square on DG; hence AF is equal to AG, and Euc. 1. 8, the angle BAC is bisected by AD. 38. The line required will be found to be equal to half the sum of the two sides of the triangle. 39. Apply Euc. I. 1, 9. 40. The angle to be trisected is one-fourth of a right angle. If an equilateral triangle be described on one of the sides of a triangle which contains the given angle, and a line be drawn to bisect that angle of the equilateral triangle which is at the given angle, the angle contained between this line and the other side of the triangle will be one-twelfth of a right angle, or equal to one-third of the given angle. It may be remarked, generally, that any angle which is the half, fourth, eighth, &c. part of a right angle, may be trisected by Plane Geometry. 41. Apply Euc. 1. 20. 42. Let ABC, DBC be two equal triangles on the same base, of which ABC is isosceles, fig. Euc. 1. 37. By producing AB and making AG equal to AB or AC, and joining GD, the perimeter of the triangle ABC may be shewn to be less than the perimeter of the triangle DBC. 43. Apply Euc. 1. 20. 44. For the first case, see Theo. 32, p. 76: for the other two cases, apply Euc. 1. 19. 45. This is obvious from Euc. 1. 26. 46. By Euc. 1. 29, 6, FC may be shewn equal to each of the lines EF, FG. 47. Join GA and AF, and prove GA and AF to be in the same straight line. 48. Let the straight line drawn through D parallel to BC meet the side AB in E, and AC in F. Then in the triangle EBD, EB is equal to ED, by Euc. 1. 29, 6. Also, in the triangle EAD, the angle EAD may be shewn equal to the angle EDA, whence EA is equal to ED, and therefore AB is bisected in E. In a similar way it may be shewn, by bisecting the angle C, that AC is bisected in F. Or the bisection of AC in F may be proved when AB is shewn to be bisected in E. 49. The triangle formed will be found to have its sides respectively parallel to the sides of the original triangle. 50. If a line equal to the given line be drawn from the point where the two lines meet, and parallel to the other given line; a parallelogram may be formed, and the construction effected. 51. Let ABC be the triangle; AD perpendicular to BC, AE drawn to the bisection of BC, and. AF bisecting the angle BAC. Produce AD and make DA' equal to AD: join FA', EA'. 52. If the point in the base be supposed to be determined, and lines drawn from it parallel to the sides, it will be found to be in the line which bisects the vertical angle of the triangle. 53. Let ABC be the triangle, at C draw CD perpendicular to CB and equal to the sum of the required lines, through D draw DE parallel to CB meeting AC in E, and draw EF parallel to DC, meeting BC in F. Then EF is equal to DC. Next produce CB, making CG equal to CE, and join EG cutting AB in H. From H draw HK perpendicular to EAC, and HL perpendicular to BC. Then HK and HL together are equal to DC. The proof depends on Theorem 27, p. 75. 54. Let C' be the intersection of the circles on the other side of the base, and join AC', BC'. Then the angles CBA, C'BA being equal, the angles CBP, C'BP are also equal, Euc. 1. 13: next by Euc. 1. 4, CP, PC are proved equal; lastly prove CC' to be equal to CP or PC'. 55. In the fig. Euc. 1. 1, produce AB both ways to meet the circles in D and E, join CD, CE, then CDE is an isosceles triangle, having each of the angles at the base one-fourth of the angle at the vertex. At E draw EG perpendicular to DB and meeting DC produced in G. Then CEG is an equilateral triangle. 56. Join CC, and shew that the angles CC'F, CC'G are equal to two right angles; also that the line FC'G is equal to the diameter. 57. Construct the figure and by Euc. 1. 32. If the angle BAC be a right angle, then the angle BDC is half a right angle. 58. Let the lines which bisect the three exterior angles of the triangle ABC form a new triangle A'B'C'. Then each of the angles at A', B', C' may be shewn to be equal to half of the angles at A and B, B and C, C and A respectively. And it will be found that half the sums of every two of three unequal numbers whose sum is constant, have less differences than the three numbers themselves. 59. The first case may be shewn by Euc. 1. 4: and the second by Euc. 1. 32, 6, 15. 60. At D any point in a line EF, draw DC perpendicular to EF and equal to the given perpendicular on the hypotenuse. With centre C and radius equal to the given base describe a circle cutting EF in B. At C draw CA perpendicular to CB and meeting EF in A. Then ABC is the triangle required. 61. Let ABC be the required triangle having the angle ACB a right angle. In BC produced, take CE equal to AC, and with center B and radius BA describe a circular arc cutting CE in D, and join AD. Then DE is the difference between the sum of the two sides AC, CB and the hypotenuse AB; also one side AC the perpendicular is given. Hence the construction. On any line EB take EC equal to the given side, ED equal to the given difference. At C, draw CA perpendicular to CB, and equal to EC, join AD, at A in AD make the angle DAB equal to ADB, and let AB meet EB in B. Then ABC is the triangle required. 62. (1) Let ABC be the triangle required, having ACB the right angle. Produce AB to D making AD equal to AC or CB: then BD is the sum of the sides. Join DC: then the angle ADC is one-fourth of a right angle, and DBC is one-half of a right angle. Hence to construct: at B in BD make the angle DBM equal to half a right angle, and at D the angle BDC equal to one-fourth of a right angle, and let DC meet BM in C. At C draw CA at right angles to BC meeting BD in A: and ABC is the triangle required. (2) Let ABC be the triangle, C the right angle: from AB cut off AD equal to AC; then BD is the difference of the hypotenuse and one side. Join CD; then the angles ACD, ADC are equal, and each is half the supplement of DAC, which is half a right angle. Hence the construction. 63. Take any straight line terminated at A. Make AB equal to the difference of the sides, and AC equal to the hypotenuse. At B make the angle CBD equal to half a right angle, and with center A and radius AC describe a circle cutting BD in D: join AD, and draw IDE perpendicular to AC. Then ADE is the required triangle. 64. Let BC the given base be bisected in D. At D draw DE at right angles to BC and equal to the sum of one side of the triangle and the perpendicular from the vertex on the base: join DB, and at B in BE make the angle EBA equal to the angle BED, and let BA meet DE in A: join AC, and ABC is the isosceles triangle. 65. This construction may be effected by means of Prob. 4, p. 71. 66. The perpendicular from the vertex on the base of an equilateral triangle bisects the angle at the vertex, which is two-thirds of one right angle. 67. Let ABC be the equilateral triangle of which a side is required to be found, having given BD, CD the lines bisecting the angles at B, C. Since the angles DBC, DCB are equal, each being one-third of a right angle, the sides BD, DC are equal, and BDC is an isosceles triangle having the angle at the vertex the supplement of a third of two right angles. Hence the side BC may be found. 68. Let the given angle be taken, (1) as the included angle between the given sides; and (2) as the opposite angle to one of the given sides. In the latter' case, an ambiguity will arise if the angle be an acute angle, and opposite to the less of the two given sides. 69. Let ABC be the required triangle, BC the given base, CD the given difference of the sides AB, AC: join BD, then DBC by Euc. 1. 18, can be shewn to be half the difference of the angles at the base, and AB is equal to AD. Hence at B in the given base BC, make the angle CBD equal to half the difference of the angles at the base. On CB take CE equal to the difference of the sides, and with center C and radius CE, describe a circle cutting BD in D: join CD and produce it to A, making DA equal to DB. Then ABC is the triangle required. 70. On the line which is equal to the perimeter of the required triangle describe a triangle having its angles equal to the given angles. Then bisect the angles at the base; and from the point where these lines meet, draw lines parallel to the sides and meeting the base. 71. Let ABC be the required triangle, BC the given base, and the side AB greater than AC. Make AD equal to AC, and draw CD. Then the angle BCD may be shewn to be equal to half the difference, and the angle DCA equal to half the sum of the angles at the base. Hence ABC, ACB the angles at the base of the triangle are known. 72. Let the two given lines meet in A, and let B be the given point. If BC, BD be supposed to be drawn making equal angles with AC, and if AD and DC be joined, BCD is the triangle required, and the figure ACBD may be shewn to be a parallelogram. Whence the construction. 73. It can be shewn that lines drawn from the angles of a triangle to bisect the opposite sides, intersect each other at a point which is twothirds of their lengths from the angular points from which they are drawn. Let ABC be the triangle required, AD, BE, CF the given lines from the angles drawn to the bisections of the opposite sides and intersecting in G. Produce GD, making DH equal to DG, and join BH, CH: the figure GBHC is a parallelogram. Hence the construction. 74. Let ABC (fig. to Euc. 1. 20.) be the required triangle, having the base BC equal to the given base, the angle ABC equal to the given angle, and the two sides BA, AC together equal to the given line BD. Join DC, then since AD is equal to AC, the triangle ACD is isosceles, and therefore the angle ADC is equal to the angle ACD. Hence the construction. 75. Let ABC be the required triangle (fig. to Euc. 1. 18), having the angle ACB equal to the given angle, and the base BC equal to the given line, also CD equal to the difference of the two sides AB, AC. If BD be joined, then ABD is an isosceles triangle. Hence the synthesis. Does this construction hold good in all cases? 76. Let ABC be the required triangle, (fig. Euc. 1. 18), of which the side BC is given and the angle BAC, also CD the difference between the sides AB, AC. Join BD; then AB is equal to AD, because CD is their difference, and the triangle ABD is isosceles, whence the angle ABD is equal to the angle ADB; and since BAD and twice the angle ABD are equal to two right angles, it follows that ABD is half the supplement of the given angle BAC. Hence the construction of the triangle. 77. Let AB be the given base: at A draw the line AD to which the line bisecting the vertical angle is to be parallel. At B draw BE parallel to AD; from A draw AE equal to the given sum of the two sides to meet BE in E. At B make the angle EBC equal to the angle BEA, and draw CF parallel to AD. Then ACB is the triangle required. 78. Take any point in the given line, and apply Euc. 1. 23, 31. 79. On one of the parallel lines take EF equal to the given line, and with center E and radius EF describe a circle cutting the other in G. Join EG, and through A draw ABC parallel to EG. 80. This will appear from Euc. 1. 29, 15, 26. 81. Let AB, AC, AD, be the three lines. Take any point E in AC, and on EC make EF equal to EA, through F draw FG parallel to AB, join GE and produce it to meet AB in H. Then GE is equal to GH. 82. Apply Euc. 1. 32, 29. 83. From E draw EG perpendicular on the base of the triangle, then ED and EF may each be proved equal to EG, and the figure shewn to be equilateral. Three of the angles of the figure are right angles. 84. The greatest parallelogram which can be constructed with given sides can be proved to be rectangular. 85. Let AB be one of the diagonals: at A in AB make the angle BAC less than the required angle, and at A in AC make the angle CAD equal to the required angle. Bisect AB in E and with center E and radius equal to half the other diagonal describe a circle cutting AC, AD in F, G. Join FB, BG: then AFBG is the parallelogram required. 86. This problem is the same as the following; having given the base of a triangle, the vertical angle and the sum of the sides, to construct the triangle. This triangle is one half of the required parallelogram. 87. Draw a line AB equal to the given diagonal, and at the point A make an angle BAC equal to the given angle. Bisect AB in D, and through D draw a line parallel to the given line and meeting AC in C. This will be the position of the other diagonal. Through B draw BE parallel to CA, meeting CD produced in E; join AE, and BC. ACBE is the parallelogram required. 88. Construct the figures and by Euc. 1. 24. 89. By Euc. 1. 4, the opposite sides may be proved to be equal. Then 90. Let ABCD be the given parallelogram; construct the other parallelogram A'B'C'D' by drawing the lines required, also the diagonals AC, A'C', and shew that the triangles ABC, A'B'C' are equiangular. 91. A'D' and B'C' may be proved to be parallel. 92. Apply Euc. 1. 29, 32. 93. The points D, D', are the intersections of the diagonals of two rectangles: if the rectangles be completed, and the lines OD, OD' be produced, they will be the other two diagonals. 94. Let the line drawn from A fall without the parallelogram, and |