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', G'. Join FG', and AFG will be the triangle. Draw DH perpendicular to FG and join GD. 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 BC. 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 AE 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 FD: 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 DEF, join Bf, and produce it to meet AC in F. Through F draw FD parallel to fd, F'E' parallel to ƒe, and join D'E', then the triangle D'E'F' 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 BCDE 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 FĞ, 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. base BC an angle equal to one of the given Draw AE parallel to BC and take AD to sides. Join BE cutting AC in F.

Draw AD making with the angles of the parallelogram. AE in the given ratio of the

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 may be proved to be a square.

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 quadrilateral 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. Byconstructing 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

may be proved to have a certain ratio, and the triangles ACB, CPM in the same way, may be proved to have the same ratio.

111. Let BA, AC be the bounding radii, and D a point in the arc of a quadrant. Bisect BAC by AE, and draw through D, the line HDGP perpendicular to AE at G, and meeting AB, AC, produced in H, P. From H draw HM to touch the circle of which BC is a quadrantal arc; produce AH, making HL equal to HM, also on HA, take HK equal to HM. Then K, L, are the points of contact of two circles through D which touch the bounding radii, AB, AC.

Join DA. Then, since BAC is a right angle, AK is equal to the radius of the circle which touches BA, BC in K, K'; and similarly, AL is the radius of the circle which touches them in L, L'. Also, HAP being an isosceles triangle, and AD drawn to the base, AD is shewn to be equal to AK. KL. Euc. III. 36; 11. 5, Cor.

112. Let E, F, G be the centers of the circles inscribed in the triangles ABC, ADB, ACD. Draw EH, FK, GL perpendiculars on BC, BA, AC respectively, and join CE, EB; BF, FA; CG, GA. Then the relation between R, r, r', or EH, FK, GL may be found from the similar triangles, and the property of right-angled triangles.

113. The two hexagons consist each of six equilateral triangles, and the ratio of the hexagons is the same as the ratio of their equilateral triangles.

114. The area of the inscribed equilateral triangle may be proved to be equal to half of the inscribed hexagon, and the circumscribed triangle equal to four times the inscribed triangle.

115. The pentagons are similar figures, and can be divided into the same number of similar triangles. Euc. vI. 19.

116. Let the sides AB, BC, CA of the equilateral triangle ABC touch the circle in the points D, E, F, respectively. Draw AE cutting the circumference in G; and take O the center of the circle and draw OD: draw also HGK touching the circle in G. The property may then be shewn by the similar triangles AHG, AOD.