59. Apply Euc. III. 27. 1. 6.

60. Let two unequal circles cut one another, and let the line ABC drawn through B, one of the points of intersection, be the line required, such that AB is equal to BC. Join O, O' the centers of the circles, and draw OP, O'P' perpendiculars on ABC, then PB is equal to BP'; through O' draw O'D parallel to PP'; then ODO' is a right-angled triangle, and a semicircle described on OO' as a diameter will pass through the point D. Hence the synthesis. If the line ABC be supposed to move round the point B and its extremities A, C to be in the extremities of the two circles, it is manifest that ABC admits of a maximum.

61. Suppose the thing done, then it will appear that the line joining the points of intersection of the two circles is bisected at right angles by the line joining the centers of the circles. Since the radii are known, the centers of the two circles may be determined.

Join

62. Let the circles intersect in A, B; and let CAD, EBF be any parallels passing through A, B and intercepted by the circles. CE, AB, DF. Then the figure CEFD may be proved to be a parallelogram. Whence CAD is equal to EBF.

63. Complete the circle whose segment is ADB; AHB being the other part. Then since the angle ACB is constant, being in a given segment, the sum of the arcs DE and AHB is constant. But AHB is given, hence ED is also given and therefore constant.

64. From A suppose ACD drawn, so that when BD, BC are joined, AD and DB shall together be double of AC and CB together. Then the angles ACD, ADB are supplementary, and hence the angles BCD, BDC are equal, and the triangle BCD is isosceles. Also the angles BCD, BDC are given, hence the triangle BDC is given in species. Again AD + DB = 2.AC + 2. BC, or CD = AC + BC.

Whence, make the triangle bde having its angles at d, c equal to that in the segment BDA; and make ca = cd cb, and join ab. At A make the angle BAD equal to bad, and AD is the line required.

65. The line drawn from the point of intersection of the two lines to the center of the given circle may be shewn to be constant, and the center of the given circle is a fixed point.

66.

This is at once obvious from Euc. III. 36.

67. This follows directly from Euc. III. 36.

68. Each of the lines CE, DF may be proved parallel to the common chord AB.

69. By constructing the figure and joining AC and AD, by Euc. III. 27, it may be proved that the line BC falls on BD.

70. By constructing the figure and applying Euc 1. 8, 4, the truth is manifest.

71. The bisecting line is a common chord to the two circles; join the other extremities of the chord and the diameter in each circle, and the angles in the two segments may be proved to be equal.

72. Apply Euc. III. 27; 1. 32, 6.

73. Draw a common tangent at C the point of contact of the circles, and prove AC and CB to be in the same straight line.

74. Let A, B, be the centers, and C the point of contact of the two circles; D, E the points of contact of the circles with the common tangent DE, and CF a tangent common to the two circles at C, meeting DF in E. Join DC, CE. Then DF, FC, FE may be shewn to be equal, and FC to be at right angles to AB.

75. The line must be drawn to the extremities of the diameters which are on opposite sides of the line joining the centers.

76. The sum of the distances of the center of the third circle from the centers of the two given circles, is equal to the sum of the radii of the given circles, which is constant.

77. Let the circles touch at C either externally or internally, and their diameters AC, BC through the point of contact will either coincide or be in the same straight line. CDE any line through C will cut off similar segments from the two circles. For joining AD, BE, the angles in the segments DAC, EBC are proved to be equal.

The remaining segments are also similar, since they contain angles which are supplementary to the angles DAC, EBC.

78. Let the line which joins the centers of the two circles be produced to meet the circumferences, and let the extremities of this line and any other line from the point of contact be joined. From the center of the larger circle draw perpendiculars on the sides of the right-angled triangle inscribed within it.

79. In general, the locus of a point in the circumference of a circle which rolls within the circumference of another, is a curve called the Hypocycloid; but to this there is one exception, in which the radius of one of the circles is double that of the other: in this case, the locus is a straight line, as may be easily shewn from the figure.

80. Let A, B be the centers of the circles. Draw AB cutting the circumferences in C, D., On AB take CE, DF each equal to the radius of the required circle: the two circles described with centers A, B, and radii AE, BF, respectively, will cut one another, and the point of intersection will be the center of the required circle.

81. Apply Euc. III. 31.

82. Apply Euc. II. 21.

83. When the tangent is on the same side of the two circles. Join C, C' their centers, and on CC' describe a semicircle. With center C' and radius equal to the difference of the radii of the two circles, describe another circle cutting the semicircle in D: join DC and produce it to meet the circumference of the given circle in B. Through C draw CA parallel to DB and join BA; this line touches the two circles.

(2) When the tangent is on the alternate sides. Having joined C, C'; on CC' describe a semicircle; with center C, and radius equal to the sum of the radii of the two circles describe another circle cutting the semicircle in D, join CD cutting the circumference in A, through C draw CB parallel to CA and join AB.

84. The possibility is obvious. The point of bisection of the segment intercepted between the convex circumferences will be the center of one of the circles: and the center of a second circle will be found to be the point of intersection of two circles described from the centers of the given circles with their radii increased by the radius of the second circle. The line passing through the centers of these two circles will be the locus of the centers of all the circles which touch the two given circles.

85. At any points P, R in the circumferences of the circles, whose centers are A, B, draw PQ, RS, tangents equal to the given lines, and join AQ, BS. These being made the sides of a triangle of which AB is the base, the vertex of the triangle is the point required.

86. In each circle draw a chord of the given length, describe circles concentric with the given circles touching these chords, and then draw a straight line touching these circles.

87. Within one of the circles draw a chord cutting off a segment equal to the given segment, and describe a concentric circle touching the chord: then draw a straight line touching this latter circle and the other given circle.

88. The tangent may intersect the line joining the centers, or the line produced. Prove that the angle in the segment of one circle is equal to the angle in the corresponding segment of the other circle.

89.

Join the centers A, B; at C the point of contact draw a tangent, and at A draw AF cutting the tangent in F, and making with CF an angle equal to one-fourth of the given angle. From F draw tangents to the circles.

90.

Let C be the center of the given circle, and D the given point in the given line AB. At D draw any line DE at right angles to AB, then the center of the circle required is in the line AE. Through C draw a diameter FG parallel to DE, the circle described passing through the points E, F, G will be the circle required.

91. Apply Euc. III. 18.

92. Let A, B, be the two given points, and C the center of the given circle. Join AC, and at C draw the diameter DCE perpendicular to AC, and through the points A, D, E describe a circle, and produce AC to meet the circumference in F. Bisect AF in G, and AB in H, and draw GK, HK, perpendiculars to AF, AB respectively and intersecting in K. Then K is the center of the circle which passes through the points A, B, and bisects the circumference of the circle whose center is C.

93. Let D be the given point and EF the given straight line. (fig. Euc. III. 32.) Draw DB to make the angle DBF equal to that contained in the alternate segment. Draw BA at right angles to EF, and DA at right angles to DB and meeting BA in A. Then AB is the diameter of the circle.

94. Let A, B be the given points, and CD the given line. From E the middle of the line AB, draw EM perpendicular to AB, meeting CD in M, and draw MA. In EM take any point F; draw FH to make the given angle with CD; and draw FG equal to FH, and meeting MA produced in G. Through A draw AP parallel to FG, and CPK parallel to FH. Then P is the center, and C the third defining point of the circle required: and AP may be proved equal to CP by means of the triangles GMF, AMP; and HMF, CMP, Euc. vI. 2. Also CPK the diameter makes with CD the angle KCD equal to FHD, that is, to the given angle.

95. Let A, B be the two given points, join AB and bisect AB in C, and draw CD perpendicular to AB, then the center of the required circle will be in CD. From O the center of the given circle draw CFG parallel to CD, and meeting the circle in F and AB produced in G. At F draw a chord FF equal to the given chord. Then the circle which passes through the points at B and F, passes also through F.

96. Let the straight line joining the centers of the two circles be produced both ways to meet the circumference of the exterior circle.

97. Let A be the common center of two circles, and BCDE the chord such that BE is double of CD. From A, B draw AF, BG perpendicular to BE. Join AC, and produce it to meet BG in G. Then AC may be shewn to be equal to CG, and the angle CBG being a right angle, is the angle in the semicircle described on CG as its diameter.

98. The lines joining the common center and the extremities of the chords of the circles, may be shewn to contain unequal angles, and the angles at the centers of the circles are double the angles at the circumferences, it follows that the segments containing these unequal angles are not similar.

99. Let AB, AC be the straight lines drawn from A, a point in

the outer circle to touch the inner circle in the points D, E, and meet the outer circle again at B, C. Join BC, DE. Prove BC double of DE. Let O be the center, and draw the common diameter AOG intersecting BC in F, and join EF. Then the figure DBFE may be proved to be a parallelogram.

100. This appears from Euc. 111. 14.

101. The given point may be either within or without the circle. Draw a chord in the circle equal to the given chord, and describe a concentric circle touching the chord, and through the given point draw a line touching this latter circle.

102. The diameter of the inner circle must not be less than one-third of the diameter of the exterior circle.

103. Suppose AD, DB to be the tangents to the circle AEB containing the given angle. Draw DC to the center C and join CA, CB. Then the triangles ACD, BCD are always equal: DC bisects the given angle at D and the angle ACB. The angles CAB, CBD, being right angles, are constant, and the angles ADC, BDC are constant, as also the angles ACD, BCD; also AC, CB the radii of the given circle. Hence the locus of D is a circle whose center is C and radius CD.

104. Let C be the center of the inner circle; draw any radius CD, at D draw a tangent CE equal to CD, join CE, and with center C and radius CE describe a circle and produce ED to meet the circle again in F.

105. Take C the center of the given circle, and draw any radius CD, at D draw DE perpendicular to DC and equal to the length of the required tangent; with center C and radius CE describe a circle.

106. This is manifest from Euc. 111. 36.

107. Let AB, AC be the sides of a triangle ABC. From A draw the perpendicular AD on the opposite side, or opposite side produced, The semicircles described on AB, BC both pass through D. Euc. 111. 31. 108. Let A be the right angle of the triangle ABC, the first property follows from the preceding Theorem 107. Let DE, DF be drawn to E, F the centers of the circles on AB, AC and join EF. Then ED may be proved to be perpendicular to the radius DF of the circle on AC at the point D.

109. Let ABC be a triangle, and let the arcs be described on the sides externally containing angles, whose sum is equal to two right angles. It is obvious that the sum of the angles in the remaining segments is equal to four right angles. These arcs may be shewn to intersect each other in one point D. Let a, b, c be the centers of the circles on BC, AC, AB. Join ab, bc, ca; Ab, bC, Ca; aB, Bc, cA; bD, cD, aD. Then the angle cba may be proved equal to one-half of the angle AbС. Similarly, the other two angles of abc.

110. It may be remarked, that generally, the mode of proof by which, in pure geometry, three lines must, under specified conditions, pass through the same point, is that by reductio ad absurdum. This will for the most part require the converse theorem to be first proved or taken for granted.

The converse theorem in this instance is, "If two perpendiculars drawn from two angles of a triangle upon the opposite sides, intersect in a point, the line drawn from the third angle through this point will be perpendicular to the third side."

The proof will be formally thus: Let EHD be the triangle, AC, BD two perpendiculars intersecting in F. If the third perpendicular EG do not pass through F, let it take some other position as EH; and through F draw EFG to meet AD in G. Then it has been proved that

EG is perpendicular to AD: whence the two angles EHG, EGH of the triangle EGH are equal to two right angles :—which is absurd.

111. The circle described on AB as a diameter will pass through E and D. Then Euc. III. 36.

112. Since all the triangles are on the same base and have equal vertical angles, these angles are in the same segment of a given circle. The lines bisecting the vertical angles may be shewn to pass through the extremity of that diameter which bisects the base.

113. Let AC be the common base of the triangles, ABC the isosceles triangle, and ADC any other triangle on the same base AC and between the same parallels AC, BD. Describe a circle about ABC, and let it cut AD in E and join EC. Then, Euc. 1. 17, III. 21.

114. Let ABC be the given isosceles triangle having the vertical angle at C, and let FG be any given line. Required to find a point P in FG such that the distance PA shall be double of PC. Divide AC in D so that AD is double of DC, produce AC to E and make AE double of AC. On DE describe a circle cutting FG in P, then PA is double of PC. This is found by shewing that AP2 = 4.PC2. 115. On any two sides of the triangle, describe segments of circles each containing an angle equal to two-thirds of a right angle, the point of intersection of the arcs within the triangle will be the point required, such that three lines drawn from it to the angles of the triangle shall contain equal angles. Euc. III. 22.

116. Let A be the base of the tower, AB its altitude, BC the height of the flagstaff, AD a horizontal line drawn from A. If a circle be described passing through the points B, C, and touching the line AD in the point E: E will be the point required. Give the analysis.

117. If the ladder be supposed to be raised in a vertical plane, the locus of the middle point may be shewn to be a quadrantal arc of which the radius is half the length of the ladder.

118. The line drawn perpendicular to the diameter from the other extremity of the tangent is parallel to the tangent drawn at the extremity of the diameter.

119. Apply Euc. III. 21.

120. Let A, B, C, be the centers of the three equal circles, and let them intersect one another in the point D: and let the circles whose centers are A, B intersect each other again in E; the circles whose centers are B, C in F; and the circles whose centers are C, A in G. Then FG is perpendicular to DE; DG to FC; and DF to GE. Since the circles are equal, and all pass through the same point D, the centers A, B, C are in a circle about D whose radius is the same as the radius of the given circles. Join AB, BC, CA; then these will be perpendicular to the chords DE, DF, DG. Again, the figures DAGC, DBFC, are equilateral, and hence FG is parallel to AB; that is, perpendicular to DE. Similarly for the other two cases.

121. Let E be the center of the circle which touches the two equal circles whose centers are A, B. Join AE, BE which pass through the points of contact F, G. Whence AE is equal to EB. Also CD the common chord bisects AB at right angles, and therefore the perpendicular from E on AB coincides with CD.

122. Let three circles touch each other at the point A, and from A let a line ABCD be drawn cutting the circumferences in B, C, D. Let O, O', O" be the centers of the circles, join BO, CO', DO", these lines are parallel to one another. Euc. 1. 5. 28.

123. Proceed as in Theorem 110, supra.