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. Then 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.

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

let CC', BB', DD', be the perpendiculars from C, B, D, on the line drawr from A; from B draw BE parallel to AC', and the truth is manifest, Next, let the line from A be drawn so as to fall within the parallelogram,

95. Let the diagonals intersect in E. In the triangles DCB, CDA, two angles in each are respectively equal and one side DE: wherefore the diagonals DB, AC are equal: also since DE, EC are equal, it follows that EA, EB are equal. Hence DEC, AEB are two isosceles triangles having their vertical angles equal, wherefore the angles at their bases are equal respectively, and therefore the angle CDB is equal to DBA.

96. (1) By supposing the point P found in the side AB of the parallelogram ABCD, such that the angle contained by AP, PC may be bisected by the line PD; CP may be proved equal to CD; hence the solution.

(2) By supposing the point P found in the side AB produced, so that PD may bisect the angle contained by ABP and PC; it may be shewn that the side AB must be produced, so that BP is equal to BD.

97. This may be shewn by Euc. 1. 35.

98. Let D, E, F be the bisections of the sides AB, BC, CA of the triangle ABC: draw DE, EF, FD; the triangle DEF is one-fourth of the triangle ABC. The triangles DBE, FBE are equal, each being one-fourth of the triangle ABC: DF is therefore parallel to BE, and DBEF is a parallelogram of which DE is a diagonal.

99. This may be proved by applying Euc 1. 38.

100. Apply Euc. 1. 37, 38.

101. On any side BC of the given triangle ABC, take BD equal to the given base; join AD, through C draw CE parallel to AD, meeting BA produced if necessary in E, join ED; then BDE is the triangle required. By a process somewhat similar the triangle may be formed when the altitude is given.

102. Apply the preceding problem (101) to make a triangle equal to one of the given triangles and of the same altitude as the other given triangle. Then the sum or difference can be readily found.

103. First construct a triangle on the given base equal to the given triangle; next form an isosceles triangle on the same base equal to this triangle.

104, Make an isosceles triangle equal to the given triangle, and then this isosceles triangle into an equal equilateral triangle.

105. Make a triangle equal to the given parallelogram upon the given line, and then a triangle equal to this triangle, having an angle equal to the given angle.

106. If the figure ABCD be one of four sides; join the opposite angles A, C of the figure, through D draw DE parallel to AC meeting BC produced in E, join AE:-the triangle ABE is equal to the foursided figure ABCD.

If the figure ABCDE be one of five sides, produce the base both ways, and the figure may be transformed into a triangle, by two constructions similar to that employed for a figure of four sides. If the figure consists of six, seven, or any number of sides, the same process must be repeated. 107. Draw two lines from the bisection of the base parallel to the two sides of the triangle.

108. This may be shewn ex absurdo.

109. On the same base AB, and on the same side of it, let two triangles ABC, ABD be constructed, having the side BD equal to BC, the angle ABC a right angle, but the angle ABD not a right angle; then the triangle ABC is greater than ABD, whether the angle ABD be acute or obtuse. 110. Let ABC be a triangle whose vertical angle is A, and whose

base BC is bisected in D; let any line EDG be drawn through D, meeting AC the greater side in G and AB produced in E, and forming a triangle AEG having the same vertical angle A. Draw BH parallel to AC, and the triangles BDH, GDC are equal. Euc. 1. 26.

111. Let two triangles be constructed on the same base with equal perimeters, of which one is isosceles. Through the vertex of that which is not isosceles draw a line parallel to the base, and intersecting the perpendicular drawn from the vertex of the isosceles triangle upon the common base. Join this point of intersection and the extremities of the base. 112. (1) DF bisects the triangle ABC (fig. Prop. 6, p. 73.) On each side of the point F in the line BC, take FG, FH, each equal to one-third of BF, the lines DG, DH shall trisect the triangle. Or,

Let ABC be any triangle, D the given point in BC. Trisect BC in E, F. Join AD, and draw EG, FH parallel to AD. Join DG, DH; these lines trisect the triangle. Draw AE, AF and the proof is manifest.

(2) Let ABC be any triangle; trisect the base BC in D, E, and join AD, AE. From D, E, draw DP, EP parallel to AB, AC and meeting in P. Join AP, BP, CP; these three lines trisect the triangle.

(3) Let P be the given point within the triangle ABC. Trisect the base BC in D, E. From the vertex A draw AD, AE, AP. Join PD, draw AG parallel to PD and join PG. Then BGPA is one-third of the triangle. The problem may be solved by trisecting either of the other two sides and making a similar construction.

113. The base may be divided into nine equal parts, and lines may be drawn from the vertex to the points of division. Or, the sides of the triangle may be trisected, and the points of trisection joined.

114. It is proved, Euc. 1. 34, that each of the diagonals of a parallelogram bisects the figure, and it may be shewn that they also bisect each other. It is hence manifest that any straight line, whatever may be its position, which bisects a parallelogram, must pass through the intersection of the diagonals.

115. See the remark on the preceding problem 114.

116. Trisect the side AB in E, F, and draw EG, FH parallel to AD or BC, meeting DC in G and H. If the given point P be in EF, the two lines drawn from P through the bisections of EG and FH will trisect the parallelogram. If P be in FB, a line from P through the bisection of FH will cut off one-third of the parallelogram, and the remaining trapezium is to be bisected by a line from P, one of its angles. If P coincide with E or F, the solution is obvious.

117. Construct a right-angled parallelogram by Euc. 1. 44, equal to the given quadrilateral figure, and from one of the angles, draw a line to meet the opposite side and equal to the base of the rectangle, and a line from the adjacent angle parallel to this line will complete the rhombus. 118. Bisect BC in D, and through the vertex A, draw AE parallel to BC, with center D and radius equal to half the sym of AB, AC, describe a circle cutting AE in E.

119. Produce one side of the square till it becomes equal to the diagonal, the line drawn from the extremity of this produced side and parallel to the adjacent side of the square, and meeting the diagonal produced, determines the point required.

120. Let fall upon the diagonal perpendiculars from the opposite angles of the parallelogram. These perpendiculars are equal, and each pair of triangles is situated on different sides of the same base and has equal al titudes.

121. One case is included in Theo. 120. The other case, when the point is in the diagonal produced, is obvious from the same principle. 122. The triangles DCF, ABF may be proved to be equal to half of the parallelogram by Euc. 1. 41.

123. Apply Euc. 1. 41, 38.

124. If a line be drawn parallel to AD through the point of intersection of the diagonal, and the line drawn through O parallel to AB; then by Euc. 1. 43, 41, the truth of the theorem is manifest.

125. It may be remarked that parallelograms, are divided into pairs of equal triangles by the diagonals, and therefore by taking the triangle ABD equal to the triangle ABC, the property may be easily shewn.

126. The triangle ABD is one half of the parallelogram ABCD, Euc. 1. 34. And the triangle DKC is one half of the parallelogram CDHG, Euc. 1. 41, also for the same reason the triangle AKB is one half of the parallelogram AHGB: therefore the two triangles DKC, AKB are together one half of the whole parallelogram ABCĎ. Hence the two triangles DKC, AKB are equal to the triangle ABD : take from these equals the equal parts which are common, therefore the triangle CKF is equal to the triangles AHK, KBD: wherefore also taking AHK from these equals, then the difference of the triangles CKF, AHK is equal to the triangle KBC: and the doubles of these are equal, or the difference of the parallelograms CFKG, AHKE is equal to twice the triangle KBD.

127. First prove that the perimeter of a square is less than the perimeter of an equal rectangle: next, that the perimeter of the rectangle is less than the perimeter of any other equal parallelogram.

128. This may be proved by shewing that the area of the isosceles triangle is greater than the area of any other triangle which has the same vertical angle, and the sum of the sides containing that angle is equal to the sum of the equal sides of the isosceles triangle.

129. Let ABC be an isosceles triangle (fig. Euc. 1. 42), AE perpendicular to the base BC, and AECG the equivalent rectangle. Then AC is greater than AE, &c.

130. Let the diagonal AC bisect the quadrilateral figure ABCD. Bisect AC in E, join BE, ED, and prove BE, ED in the same straight line and equal to one another.

131. Apply Euc. 1. 15.

132. Apply Euc. 1. 20.

133. This may be shewn by Euc. 1. 20.

134. Let AB be the longest and CD the shortest side of the rectangular figure. Produce AD, BC to meet in E. Then by Euc. 1. 32.

135. Let ABCD be the quadrilateral figure, and E, F, two points in the opposite sides AB, CD, join EF and bisect it in G; and through G draw a straight line HGK terminated by the sides AD, BC; and bisected in the point G. Then EF, HK are the diagonals of the required parallelogram.

136. After constructing the figure, the proof offers no difficulty.

137. If any line be assumed as a diagonal, if the four given lines taken two and two be always greater than this diagonal, a four-sided figure may be constructed having the assumed line as one of its diagonals: and it may be shewn that when the quadrilateral is possible, the sum of every three given sides is greater than the fourth.

138. Draw the two diagonals, then four triangles are formed, two on one side of each diagonal. Then two of the lines drawn through the points of bisection of two sides may be proved parallel to one diagonal, and tw

parallel to the other diagonal, in the same way as Theo. 97, supra. The other property is manifest from the relation of the areas of the triangles made by the lines drawn through the bisections of the sides.

139. It is sufficient to suggest, that triangles on equal bases, and of equal altitudes, are equal.

140. Let the side AB be parallel to CD, and let AB be bisected in E and CD in F, and let EF be drawn. Join AF, BF, then Euc. 1. 38.

141. Let BCED be a trapezium of which DC, BE are the diagonals intersecting each other in G. If the triangle DBG be equal to the triangle EGC, the side DE may be proved parallel to the side BC, by Euc. 1. 39. 142. Let ABCD be the quadrilateral figure having the sides AB, CD, parallel to one another, and AD, BC equal. Through B draw BE parallel to AD, then ABED is a parallelogram.

143. Let ABCD be the quadrilateral having the side AB parallel to CD. Let E, F be the points of bisection of the diagonals BD, AC, and join EF and produce it to meet the sides AD, BC in G and H. Through H draw LHK parallel to DA meeting DC in L and AB produced in K. Then BK is half the difference of DC and AB.

144. (1) Reduce the trapezium ABCD to a triangle BAE by Prob. 106, supra, and bisect the triangle BAE by a line AF from the vertex. If F fall without BC, through F draw FG parallel to AC or DE, and join AG.

Or thus. Draw the diagonals AC, BD: bisect BD in E, and join AE, EC. Draw FEG parallel to AC the other diagonal, meeting AD in F, and DC in G. AG being joined, bisects the trapezium,

(2) Let E be the given point in the side AD. Join EB. Bisect the quadrilateral EBCD by EF. Make the triangle EFG equal to the triangle EAB, on the same side of EF as the triangle AB. Bisect the triangle EFG by EH. EH bisects the figure.

145. If a straight line be drawn from the given point through the intersection of the diagonals and meeting the opposite side of the square; the problem is then reduced to the bisection of a trapezium by a line drawn from one of its angles.

146. If the four sides of the figure be of different lengths, the truth of the theorem may be shewn. If, however, two adjacent sides of the figure be equal to one another, as also the other two, the lines drawn from the angles to the bisection of the longer diagonal, will be found to divide the trapezium into four triangles which are equal in area to one another. Euc. 1. 38.

147. Apply Euc. 1. 47, observing that the shortest side is one half of the longest.

148. Find by Euc. 1. 47, a line the square on which shall be seven times the square on the given line. Then the triangle which has these two lines containing the right angle shall be the triangle required.

149. Apply Euc. 1. 47.

150. Let the base BC be bisected in D, and DE be drawn perpendicular to the hypotenuse AC. Join AD: then Euc. 1. 47.

151. Construct the figure, and the truth is obvious from Euc. 1. 47. 152. See Theo. 32, p. 76, and apply Euc. 1. 47.

153. Draw the lines required and apply Euc. I. 47.

154. Apply Euc. 1. 47.

155. Apply Euc. 1. 47.

156. Apply Euc. 1. 47, observing that the square on any line is four times the square on half the line.