4. Right lines from any point in the diagonal of a parallelogram to the angular points through which the diagonal does not pass, and the diagonal, divide the parallelogram into four triangles which are equal, two by two. 5. If one diagonal of a quadrilateral bisects the other, it also bisects the quadrilateral, and conversely. 6. If two △s ABC, ABD be on the same base AB, and between the same parallels, and if a parallel to AB meet the sides AC, BC in the point E, F; and the sides AD, BD in the point G, H; then EF = GH. 7. If instead of triangles on the same base we have triangles on equal bases and between the same parallels, the intercepts made by the sides of the triangles on any parallel to the bases are equal. 8. If the middle points of any two sides of a triangle be joined, the triangle so formed with the two half sides is one-fourth of the whole. 9. The triangle whose vertices are the middle points of two sides, and any point in the base of another triangle, is one-fourth of that triangle. 10. Bisect a given triangle by a right line drawn from a given point in one of the sides. 11. Trisect a given triangle by three right lines drawn from a given point within it. 12. Prove that any right line through the intersection of the diagonals of a parallelogram bisects the parallelogram. 13. The triangle formed by joining the middle point of one of the non-parallel sides of a trapezium to the extremities of the opposite side is equal to half the trapezium. PROP. XXXIX.—Theorem. Equal triangles (BAC, BDC) on the same base (BC) and on the same side of it are between the same parallels. Dem.—Join AD. Then if AD be not parallel to BC, let AE be parallel to it, and let it cut BD in E. Join EC. Now since the triangles BEC, BAC are on the same base BC, and between the same parallels BC, AE, they are equal [xxxvii.]; but the triangle BAC is equal to the triangle BDC (hyp.). Therefore (Axiom i.) the triangle BEC is equal to the triangle BDC—that is, a part equal to the whole which is absurd. Hence AD must be parallel to BC. PROP. XL.—Theorem. Equal triangles (ABC, DEF) on equal bases (BC, EF) which form parts of the same right line, and on the same side of the line, are between the same parallels. Dem.—Join AD. If AD be not parallel to BF, let AG be parallel to it. Join GF. Now since the triangles GEF and ABC are on equal bases BC, EF, and between the same parallels BF, AG, they are equal [xxxviii.]; but the triangle DEF is equal to the triangle ABC (hyp.). Hence GEF is equal to DEF (Axiom i.)—that is, a part equal to the whole, which is absurd. Therefore AD must be parallel to BF. Def.—The altitude of a triangle is the perpendicular from the vertex on the base. Exercises. 1. Triangles and parallelograms of equal bases and altitudes are respectively equal.