to be equal to DF and to DP. When P coincides with Q, any point D in CD fulfils the conditions of the problem ; that is, there are innumerable solutions. 62. It may be proved that the vertices of the two triangles which are similar in the same segment of a circle, are in the extremities of a chord parallel to the chord of the given segment. 63. For let the circle be described about the triangle EAC, then by the converse to Euc. 111. 32 ; the truth of the proposition is manifest. 64. Let the figure be constructed, and the similarity of the two triangles will be at once obvious from Euc iii. 32.; Euc. 1. 29. 65. In the arc AB (fig. Euc. iv. 2) let any point K be taken, and from K let KL, KM, KN be drawn perpendicular to AB, AC, BC respectively, produced if necessary, also let LM, LN be joined, then MLN may be shewn to be a straight line. Draw AK, BK, CK, and by Euc. III. 31, 22, 21; Euc. 1. 14. 66. Let AB a chord in a circle be bisected in C, and DE, FG two chords drawn through C; also let their extremities DG, FE be joined intersecting CB in H, and AC in K; then AK is equal to HB. Through H draw MÜL parallel to EF meeting FG in M, and DE produced in L. Then by means of the equiangular triangles, HC may be proved to be equal to CK, and hence AK is equal to HB. 67. Let A, B be the two given points, and let P be a point in the locus so that PA, PB being joined, PA is to PB in the given ratio. Join AB and divide it in C in the given ratio, and join PC. Then PC bisects the angle APB. Euc. vi. 3. Again, in AB produced, take AD to AB in the given ratio, join PD and produce AP to E, then PD bisects the angle BPE. Euc. vi. A. Whence CPD is a right angle, and the point P lies in the circumference of a circle whose diameter is CD. 68. Let ABC be a triangle, and let the line AD bisecting the vertical angle A be divided in E, so that BC: BA+AC :: AE: EĎ. By Euc. vi. 3, may be deduced BC : BA+AC:: AC: AD. Whence may be proved that CE bisects the angle ACD, and by Euc. iv. 4, that E is the center of the inscribed circle. 69. By means of Euc. iv. 4, and Euc. vi. C. this theorem may be shewn to be true. 70. Divide the given base BC in D, so that BD may be to DC in the ratio of the sides. At B, D draw BB', DD' perpendicular to BC and equal to BD, DC respectively. Join B'D' and produce it to meet BC produced in O. With center and radius OD, describe a circle. From A any point in the circumference join AB, AC, AO. Prove that AB is to AC as BD to DC. Or thus. If ABC be one of the triangles. Divide the base BC in D so that BA is to AC as BD to DC. Produce BC and take DO to OC as BA to AC: then O is the center of the circle. 71. Let ABC be any triangle, and from A, B let the perpendiculars AD, BE on the opposite sides intersect in P: and let AF, BG drawn to F, G the bisections of the opposite sides, intersect in Q. Also let FR, GR be drawn perpendicular to BC, AC, and meet in R: then R is the center of the circumscribed circle. Join PQ, QR; these are in the same line. Join FG, and by the equiangular triangles, GRF, APB, AP is proved double of FR. And AQ is double of QF, and the alternate angles PAQ, QFR are equal. Hence the triangles APQ, RFQ are equiangular. *72. Let C, C be the centers of the two circles, and let CC the line joining the centers intersect the common tangent PP' in T. Let the line joining the centers cut the circles in D, Q', and let PQ, P'Q' be joined; then PQ is parallel to P'Q. Join CP, C'P', and then the angle QPT may be proved to be equal to the alternate angle Q'P'T. 73. Let ABC be the triangle, and BC its base; let the circles AFB, AFC be described intersecting the base in the point F, and their diameters AD, AE, be drawn; then DA: AE :: BA : AC. For join DB, DF, EF, EC, the triangles DAB, EAC may be proved to be similar. 74. If the extremities of the diameters of the two circles be joined by two straight lines, these lines may be proved to intersect at the point of contact of the two circles; and the two right-angled triangles thus formed may be shewn to be similar by Euc. 111. 34. 75. This follows directly from the similar triangles. 76. Let the figure be constructed as in Theorem 4, p. 162, the triangle EAD being right-angled at A, and let the circle inscribed in the triangle ADE touch AD, AE, DE in the points K, L, M respectively. Then AK is equal to AL, each being equal to the radius of the inscribed circle. Also AB is equal to GC, and AB is half the perimeter of the tri. angle AED Also if GA be joined, the triangle ADE is obviously equal to the difference of AGDE and the triangle GDE, and this difference may be proved equal to the rectangle contained by the radii of the other two circles. 77. From the centers of the two circles let straight lines be drawn to the extremities of the sides which are opposite to the right angles in each triangle, and to the points where the circles touch these sides. Euc. VI. 4. 78. Let A, B be the two given points, and C a point in the circumference of the given circle. Let a circle be described through the points A, B, C and cutting the circle in another point D. Join CD, AB, and produce them to meet in E. Let EF be drawn touching the given circle in F; the circle described through the points A, B, F, will be the circle required. Joining AD and CB, by Euc. 11. 21, the triangles CEB, AED are equiangular, and by Euc. VI. 4, 16, 111. 36, 37, the given circle and the required circle each touch the line EF in the same point, and therefore touch one another. When does this solution fail ? Various cases will arise according to the relative position of the two points and the circle. 79. Let A be the given point, BC the given straight line, and D the center of the given circle. Through D draw CD perpendicular to BC, meeting the circumference in E, F. Join AF, and take FG to the diameter FE, as FC is to FA. The circle described passing through the two points A, G and touching the line BC in B is the circle required. Let it be the center of this circle ; join HB, and BF cutting the circumference of the given circle in K, and join EK. Then the tri. angles FBC, FKE being equiangular, by Euc. vi. 4, 16, and the construction, K is proved to be a point in the circumference of the circle passing through the points A, G, B. And if DK, KH be joined, DKH may be proved to be a straight line:– the straight line which joins the centers of the two circles, and passes through a common point in their circumferences. 80. Let A be the given point, B, C the centers of the two given circles. Let a line drawn through B, C meet the circumferences of the circles in G, F; E, D, respectively. In GD produced, take the point H, so that BH is to CH as the radius of the circle whose center is B to the radius of the circle whose center is C. Join AH, and take KH to DH as GH to AH. Through A, K describe a circle ALK touching the circle whose center is B, in L. Then M may be proved to be a point in the circumference of the circle whose center is C. For by joining HL and producing it to meet the circumference of the circle whose center is B in N; and joining BN, BL, and drawing CO parallel to BL, and CM parallel to BN, the line HN is proved to cut the circumference of the circle whose center is B in M, O; and CO, CM are radii. By joining GI, DM, M may be proved to be a point in the circumference of the circle ALK. And by producing BL, CM to meet in P, P is proved to be the center of ALK, and BP joining the centers of the two circles passes through L the point of contact. Hence also is shewn that PMC passes through M, the point where the circles whose centers are P and C touch each other. Note. If the given point be in the circumference of one of the circles, the construction may be more simply effected thus: Let A be in the circumference of the circle whose center is B. Join BA, and in AB produced, if necessary, take AD equal to the radius of the circle whose center is C; join DC, and at C make the angle DCE equal to the angle CDE, the point E determined by the intersection of DA pro. duced and CE, is the center of the circle. 81. Let AB, AC be the given lines and P the given point. Then if O be the center of the required circle touching AB, AC, in R, S, the line AO will bisect the given angle BAC. Let the tangent from P meet the circle in Q, and draw OQ, OS, OP, AP. Then there are given AP and the angle OAP. Also since OQP is a right angle, we have OP-QO? =OP: _OS?=PQ’ a given magnitude. Moreover the right-angled tri. angle AOS is given in species, or OS to 0A is a given ratio. Whence in the triangle AOP there is given, the angle AOP, the side AP, and the excess of OP? above the square of a line having a given ratio to OA, to determine OA. Whence the construction is obvious. 82. Let the two given lines AB, BD meet in B, and let C be the cen. ter of the given circle, and let the required circle touch the line AB, and have its center in BD. Draw CFE perpendicular to HB intersecting the circumference of the given circle in F, and produce CE, making EF equal to the radius CF. Through G draw ĠK parallel to AB, and meeting DB in K. Join CK, and through B, draw BL parallel to KC, meeting the circumference of the circle whose center is C in L; join CL and produce CL to meet BD in 0. Then ( is the center of the circle required. Draw OM perpendicular to AB, and produce EC to meet BD in N. Then by the similar triangles, OL may be proved equal to OM. 83. (1) In every right-angled triangle when its three sides are in Arithmetical progression, they may be shewn to be as the numbers 5, 4, 3. On the given line AC describe a triangle having its sides AC, AD, DC in this proportion, bisect the angles at A, C by AE, CE meeting in E, and through E draw EF, EG parallel to AD, DC meeting in F and G.. (2) Let AC be the sum of the sides of the triangle, fig. Euc. vi, 13. Upon AC describe a triangle ADC whose sides shall be in continued proportion. Bisect the angles at A and C by two lines meeting in E. \From E draw EF, EG parallel to DA, DC respectively. 84. Describe a circle with any radius, and draw within it the straight line MN cutting off a segment containing an angle equal to the given angle, Euc. III. 34. Divide MN in the given ratio in P, and at P draw PA perpendicular to MN and meeting the circumference in A. Join AM, AN, and on AP or AP produced, take AD equal to the given perpendicular, and through D draw BC parallel to MN meeting AM, AN, or these lines produced. Then ABC shall be the triangle required. 85. Let PAQ be the given angle, bisect the angle A by AB, in AB find D the center of the inscribed circle, and draw DC perpendicular to AP. In DB take DE such that the rectangle DE, DC is equal to the given rectangle. Describe a circle on DE as diameter meeting AP in F, G; and “AQ in F, Gʻ. Join FG', and AFG will be the triangle. Draw DH perpendicular to FG and join GʻD. By Euc. vi. C, the rectangle FD, DG’ is equal to the rectangle ED, DK or CD, DE. 86. On any base BC describe a segment of a circle BAC containing an angle equal to the given angle. From D the middle point of BC draw DA to make the given angle ADC with the base. Produce AD to E so that AE is equal to the given bisecting line, and through E draw FG parallel to BĈ. Join AB, AC and produce them to meet FG in F and G. 87. Employ Theorem 70, p. 310, and the construction becomes obvious. 88. Let AB be the given base, ACB the segment containing the vertical angle; draw the diameter AB of the circle, and divide it in E, in the given ratio ; on AE as a diameter, describe a circle AFE; and with center B and a radius equal to the given line, describe a circle cutting AFE in F. Then AF being drawn and produced to meet the circumscribing circle in C, and CB being joined, ABC is the triangle required. For AF is to FC in the given ratio. 89. The line CD is not necessarily parallel to AB. Divide the base AB in C, so that AC is to CB in the ratio of the sides of the triangle. Then if a point E in CD can be determined such that when AE, CE, EB, are joined, the angle AEB is bisected by CE, the problem is solved. 90. Let ABC be any triangle having the base BC. On the same base describe an isosceles triangle DBC equal to the given triangle. Bisect BC in E, and join DE, also upon BC describe an equilateral triangle. On FD, FB, take EG to EH as EF to FB: also take EK equal to EH and join GH, GK; then GHK is an equilateral triangle equal to the triangle ABC. 91. Let ABC be the required triangle, BC the hypotenuse, and FHKG the inscribed square : the side HK being on BC. Then BC may be proved to be divided in H and K, so that HK is a mean proportional between BH and KC. 92. Let ABC be the given triangle. On BC take BD equal to one of the given lines, through A draw XE parallel to BC. From B draw BE to meet AE in E, and such that BE is a fourth proportional to BC, BD, and the other given line. Join EC, produce BE to F, making BF equal to the other given line, and join FÒ: then FBD is the triangle required. 93. By means of Euc. VI. C, the ratio of the diagonals AC to BD may be found to be as AB. AD + BC.CD to AB.BE + AD.DC, figure, Euc. vi. D. 94. This property follows directly from Euc. vi. C. 95. Let ABC be any triangle, and DEF the given triangle to which the inscribed triangle is required to be similar. Draw any line de terminated by AB, AC, and on de towards AC describe the triangle def similar to DĖF, join Bf, and produce it to meet AC in F'. Through F draw F'D' parallel to fd, F'E' parallel to fe, and join D'E', then the triangle D'EF is similar to DEF. 96. The square inscribed in a right-angled triangle which has one of its sides coinciding with the hypotenuse, may be shewn to be less than that which has two of its sides coinciding with the base and perpendicular. 97. Let BCD E be the square on the side BC of the isosceles triangle ABC. Then by Euc. vi. 2, FG is proved parallel to ED or BC. 98. Let AB be the base of the segment ABD, fig. Euc. 111. 30. Bisect AB in C, take any point E in AC and make CF equal to CE: upon EF describe a square EFGH: from C draw CG and produce it to meet the arc of the segment in K. 99. Take two points on the radii equidistant from the center, and on the line joining these points, describe a square; the lines drawn from the center through the opposite angles of the square to meet the circular arc, will determine two points of the square inscribed in the sector. 100. Let ABCDE be the given pentagon. On AB, AE take equal distances AF, AG, join FG, and on FG describe a square FGKH. Join AH and produce it to meet a side of the pentagon in L. Draw LM parallel to FH meeting AE in M. Then LM is a side of the inscribed square. 101. Let ABC be the given triangle. Draw AD making with the base BC an angle equal to one of the given angles of the parallelogram. Draw AE parallel to BC and take AD to AE in the given ratio of the sides. Join BE cutting AC in F. 102. The locus of the intersections of the diagonals of all the rectangles inscribed in a scalene triangle, is a straight line drawn from the bisection of the base to the bisection of the shorter side of the triangle, 103. This parallelogram is one half of the square in the circle. 104. Analysis. Let ABCD be the given rectangle, and EFGH that to be constructed. Then the diagonals of EFGH are equal and bisect each other in P the center of the given rectangle. About EPF describe a circle meeting BD in K, and join KE, KF. Then since the rectangle EFGH is given in species, the angle EPF formed by its diagonals is given; and hence also the opposite angle EKF of the inscribed quadri. lateral PEKF is given. Also since KP bisects that angle, the angle PKE is given, and its supplement BKE is given. And in the same way, KF is parallel to another given line; and hence EF is parallel to a third given line. Again, the angle EPF of the isosceles triangle EPF is given; and hence the quadrilateral EPFK is given in species. 105. In the figure Euc. III, 30; from C draw CE, CF making with CD, the angles DCE, DCF each equal to the angle CDA or CDB, and meeting the arc ADB in E and F. Join EF, the segment of the circle described upon EF and which passes through C, will be similar to ADB. 106. The square inscribed in the circle may be shewn to be equal to twice the square on the radius; and five times the square inscribed in the semicircle to four times the square on the radius. 107. The three triangles formed by three sides of the square with segments of the sides of the given triangle, may be proved to be similar. Whence by Euc. vi. 4, the truth of the property 108. By constructing the figure, it may be shewn that twice the square inscribed in the quadrant is equal to the square on the radius, and that five times the square inscribed in the semicircle is equal to four times the square on the radius. Whence it follows that, &c. 109. By Euc. 1. 47, and Euc. VI. 4, it may be shewn, that four times the square on the radius is equal to fifteen times the square on one of the equal sides of the triangle. 110. Constructing the figure, the right-angled triangles SCT, ACB |