98. Bisect a parallelogram by a straight line drawn through a given point within it.
99. Construct a rhombus equal to a given parallelogram.
100. If two triangles have two sides of the one equal to two sides of the other, each to each, and the sum of the two angles contained by these sides equal to two right angles, the triangles are equal in area.
101. A straight line is drawn bisecting a parallelogram ADCD and meeting AD at E and BC at F: shew that the triangles EBF and CED are equal.
102. Shew that the four triangles into which a parallelogram is divided by its diagonals are equal in area.
103. Two straight lines AB and CD intersect at E, and the triangle AEC is equal to the triangle BED: shew that BC is parallel to AD.
104. ABCD is a parallelogram; from any point P in the diagonal BD the straight lines PA, PC are drawn. Shew that the triangles PAB and PCB are equal.
105. If a triangle is described having two of its sides equal to the diagonals of any quadrilateral, and the included angle equal to either of the angles between these diagonals, then the area of the triangle is equal to the area of the quadrilateral.
106. The straight line which joins the middle points of two sides of any triangle is parallel to the base.
107. Straight lines joining the middle points of adjacent sides of a quadrilateral form a parallelogram.
108. D, E are the middle points of the sides AB, AC of a triangle, and CD, BE intersect at F: shew that the triangle BFC is, equal to the quadrilateral ADFE.
109. The straight line which bisects two sides of any triangle is half the base.
110. In the base AC of a triangle take any point D; bisect AD, DC, AB, BC at the points E, F, G, H respectively: shew that EG is equal and parallel to FH.
111. Given the middle points of the sides of a triangle, construct the triangle.
112. If the middle points of any two sides of a triangle be joined, the triangle so cut off is one quarter of the whole.
113. The sides AB, AC of a given triangle ABC are bisected at the points E,F; a perpendicular is drawn from A to the opposite side, meeting it at D. Shew that the
