130. The perimeter of a quadrilateral is greater than the sum, and less than twice the sum, of the diagonals.

131. If a triangle and a quadrilateral stand on the same base, and the one figure fall within the other, that which has the greater surface shall have the greater perimeter.

132. If one angle of a triangle be equal to the sum of the other two, the triangle can be divided into two isosceles triangles.

133. If any sect joining two parallels be bisected, this point will bisect any other sect drawn through it and terminated by the parallels.

134. Through a given point draw a line such that the part intercepted between two given parallels may equal a given sect.

135. The medial from vertex to base of a triangle bisects the intercept on every parallel to the base.

136. Show that the surface of a quadrilateral equals the surface of a triangle which has two of its sides equal to the diagonals of the quadrilateral, and the included angle equal to either of the angles at which the diagonals intersect.

137. Describe a square, having given a diagonal.

138. ABC is a right-angled triangle; BCED is the square on the hypothenuse; ACKH and ABFG are the squares on the other sides. Find the center of the square ABFG (which may be done by drawing the two diagonals), and through it draw two lines, one parallel to BC, and the other perpendicular to BC. This divides the square ABFG into four congruent quadrilaterals. Through each mid-point of the sides of the square BCED draw a parallel to AB or AC. If each be extended until it meets the second of the other pair, they will cut the square BCED into a square and four quadrilaterals congruent to ACKH and

the four quadrilaterals in ABFG.

139. The orthocenter, the centroid, and the circumcenter of a triangle are collinear, and the sect between the first two is double of the sect between the last two.

140. The perpendicular from the circumcenter to any side of a triangle is half the sect from the opposite vertex to the ortho

center.

141. Sects drawn from a given point to a given circle are bisected ; find the locus of their mid-points.

142. The intersection of the lines joining the mid-points of opposite sides of a quadrilateral is the mid-point of the sect joining the mid-points of the diagonals.

143. A parallelogram has central symmetry.

SYMMETRY.

144. No triangle can have a center of symmetry, and every axis of symmetry is a medial.

145. Of two sides of a triangle, that is the greater which is cut by the perpendicular bisector of the third side.

146. If a right-angled triangle is symmetrical, the axis bisects the right angle.

147. An angle in a triangle will be acute, right, or obtuse, according as the medial through its vertex is greater than, equal to, or less than, half the opposite side.

148. If a quadrilateral has axial symmetry, the number of vertices not on the axis must be even; if none, it is a symmetrical trapezoid; if two, it is a kite.

149. A kite has the following seven properties; from each prove all the others by proving that a quadrilateral possessing it is a kite.

(1) One diagonal, the axis, is the perpendicular bisector of the other, which will be called the transverse axis.

(2) The axis bisects the angles at the vertices which it joins. (3) The angles at the end-points of the transverse axis are equal, and equally divided by the latter.

(4) Adjacent sides which meet on the axis are equal.

(5) The axis divides the kite into two triangles which are congruent, with equal sides adjacent.

(6) The transverse axis divides the kite into two triangles, each of which is symmetrical.

(7) The lines joining the mid-points of opposite sides meet on the axis, and are equally inclined to it.

150. A symmetrical trapezoid has the following five properties; from each prove all the others by proving that a quadrilateral possessing it is a symmetrical trapezoid.

(1) Two opposite sides are parallel, and have a common perpendicular bisector.

(2) The other two opposite sides are equal, and equally inclined to

either of the other sides.

(3) Each angle is equal to one, and supplemental to the other, of its two adjacent angles.

(4) The diagonals are equal, and divide each other equally.

(5) One median line bisects the angle between those sides produced which it does not bisect, and likewise bisects the angle between the two diagonals.

151. Prove the properties of the parallelogram from its central symmetry.

152. A kite with a center is a rhombus ; a symmetrical trapezoid with a center is a rectangle; if both a rhombus and a rectangle, it is a

square.

BOOK II.

153. The perpendicular from the centroid to a line outside the triangle equals one-third the sum of the perpendiculars to that line from the vertices.

154. If two sects be each divided internally into any number of parts, the rectangle contained by the two sects is equivalent to the sum of the rectangles contained by all the parts of the one, taken separately, with all the parts of the other.

155. The square on the sum of two sects is equivalent to the sum of the two rectangles contained by the sum and each of the sects.

156. The square on any sect is equivalent to four times the square on half the sect.

157. The rectangle contained by two internal segments of a sect grows less as the point of section moves from the mid-point.

158. The sum of the squares on the two segments of a sect is least when they are equal.

159. If the hypothenuse of an isosceles right-angled triangle be divided into internal or external segments, the sum of their

squares

is

equivalent to twice the square on the sect joining the point of section to the right angle.

160. Describe a rectangle equivalent to a given square, and having one of its sides equal to a given sect.

161. Find the locus of the vertices of all triangles on the same base, having the sum of the squares of their sides constant.

162. The center of a fixed circle is the point of intersection of the diagonals of a parallelogram; prove that the sum of the squares on the sects drawn from any point on the circle to the four vertices of the parallelogram is constant.

163. Thrice the sum of the squares on the sides of any pentagon is equivalent to the sum of the squares on the diagonals, together with four times the sum of the squares on the five sects joining, in order, the midpoints of those diagonals.

164. The sum of the squares on the sides of a triangle is less than twice the sum of the rectangles contained by every two of the sides.

165. If from the hypothenuse of a right-angled triangle sects be cut off equal to the adjacent sides, the square of the middle sect thus formed is equivalent to twice the rectangle contained by the extreme

sects.

BOOK III.

166. From a point, two equal sects are drawn to a circle. Prove that the bisector of their angle contains the center of the circle.

167. Describe a circle of given radius to pass through a given point and have its center in a given line.

168. Equal chords in a circle are all tangent to a concentric circle. 169. Two concentric circles intercept equal sects on any common

secant.

170. If two equal chords intersect either within or without a circle, the segments of the one equal the segments of the other.

171. Divide a circle into two segments such that the angle in the one shall be seven times the angle in the other.

172. ABC and ABC' are two triangles such that AC = AC'; prove the circle through A, B, C, equal to that through A, B, C'.

173. Pass a circle through four given points. When is the problem possible?

174. If AC and BD be two equal arcs in a circle ABCD, prove chord AB parallel to chord CD.

175. Circles described on any two sides of a triangle as diameters intersect on the third side, or the third side produced.

176. If one circle be described on the radius of another as diameter, any chord of the larger from the point of contact is bisected by the smaller.

177. If two circles cut, and from one of the points of intersection two diameters be drawn, their extremities and the other point of intersection will be collinear.

178. Find a point inside a triangle at which the three sides shall subtend equal angles.

179. On the produced altitudes of a triangle the sects between the orthocenter and the circumscribed circle are bisected by the sides of the triangle.

180. If on the three sides of any triangle equilateral triangles be described outwardly, the sects joining their circumcenters form an equilateral triangle.

181. If two chords intersect at right angles, the sum of the squares on their four segments is equivalent to the square on the diameter.

182. The opposite sides of an inscribed quadrilateral are produced to meet; prove that the bisectors of the two angles thus formed are at right angles.

183. The feet of the perpendiculars drawn from any point in its circumscribed circle to the sides of a triangle are collinear.

184. From a point P outside a circle, two secants, PAB, PDC, are drawn to the circle ABCD; AC, BD, are joined, and intersect at C. Prove that lies on the chord of contact of the tangents drawn from to the circle. Hence devise a method of drawing tangents to a circle from an external point by means of a ruler only.

185. A variable chord of a given circle passes through a fixed point; find the locus of its mid-point.

186. Find the locus of points from which tangents to a given circle contain a given angle.