2. The right line joining the middle points of two sides of a triangle is parallel to the third; for the medians from the extremities of the base to these points will each bisect the original triangle. Hence the two triangles whose base is the third side and whose vertices are the points of bisection are equal. 3. The parallel to any side of a triangle through the middle point of another bisects the third. 4. The lines of connexion of the middle points of the sides of a triangle divide it into four congruent triangles. 5. The line of connexion of the middle points of two sides of a triangle is equal to half the third side. 6. The middle points of the four sides of a convex quadrilateral, taken in order, are the angular points of a parallelogram whose area is equal to half the area of the quadrilateral. 7. The sum of the two parallel sides of a trapezium is double the line joining the middle points of the two remaining sides. 8. The parallelogram formed by the line of connexion of the middle points of two sides of a triangle, and any pair of parallels drawn through the same points to meet the third side, is equal to half the triangle. 9. The right line joining the middle points of opposite sides of a quadrilateral, and the right line joining the middle points of its diagonals, are concurrent. PROP. XLI.—Theorem. If a parallelogram (ABCD) and a triangle (EBC) be on the same base (BC) and between the same parallels, the parallelogram is double of the triangle. Dem.—Join AC. The parallelogram ABCD is double of the triangle ABC [xxxiv.]; but the triangle ABC is equal to the triangle EBC [xxxvii.]. Therefore the parallelogram ABCD is double of the triangle EBC. Cor. 1.—If a triangle and a parallelogram have equal altitudes, and if the base of the triangle be double of the base of the parallelogram, the areas are equal. Cor. 2.—The sum of the triangles whose bases are two opposite sides of a parallelogram, and which have any point between these sides as a common vertex, is equal to half the parallelogram. PROP. XLII.—Problem. To construct a parallelogram equal to a given triangle (ABC), and having an angle equal to a given angle (D). Sol.—Bisect AB in E. Join EC. Make the angle BEF [xxiii.] equal to D. Draw CG parallel to AB [xxxi.], and BG parallel to EF. EG is a parallelogram fulfilling the required conditions. Dem.—Because AE is equal to EB (const.), the triangle AEC is equal to the triangle EBC [xxxviii.], therefore the triangle ABC is double of the triangle EBC; but the parallelogram EG is also double of the triangle EBC [xli.], because they are on the same base EB, and between the same parallels EB and CG. Therefore the parallelogram EG is equal to the triangle ABC, and it has (const.) the angle BEF equal to D. Hence EG is a parallelogram fulfilling the required conditions. PROP. XLIII.—Theorem.