2∘. When it is any parallelopiped, ABCD–EFGH, the diagonal plane bisects it.
Dem.—Through the vertices A, E let planes be drawn perpendicular to the edges and cutting them in the points I, J, K; L, M, N, respectively. Then [I. xxxiv.] we have IL = BF, because each is equal to AE. Hence IB = LF. In like manner JC = MG. Hence the pyramid A–IJCB agrees in everything but position with E–LMGF; hence it is equal to it in volume. To each add the solid ABC–LME, and we have the prism AIJ–ELM equal to the prism ABC–EFG. In like manner AJK–EMN = ACD–EGH; but (1∘) the prism AIJ–ELM = AJK–EMN. Hence ABC–EFG = ACD–EGH. Therefore the diagonal plane bisects the parallelopiped.
Cor. 1.—The volume of a triangular prism is equal to the product of its base and altitude; because it is half of a parallelopiped, which has a double base and equal altitude.
Cor. 2.—The volume of any prism is equal to the product of its base and altitude; because it can be divided into triangular prisms.
PROP. IV.—Theorem.
If a pyramid (O–ABCDE) be cut by any plane (abcde) parallel to the base, the section is similar to the base.
Dem.—Because the plane AOB cuts the parallel planes ABCDE, abcde, the sections AB, ab are parallel [XI. xvi.] In like manner BC, bc are parallel. Hence the angle ABC = abc [XI. x.]. In like manner the remaining angles of the polygon ABCDE are equal to the corresponding angles of abcde. Again, because ab is parallel to AB, the triangles ABO, abO are equiangular.
| Hence | AB | : BO :: ab : bO. [VI. iv.] | |||||||||
| In like manner | BO | : BC :: bO : bc; | |||||||||
| therefore | AB | : BC :: ab : bc. [Ex æquali.] | |||||||||
| In like manner | BC | : CD :: bc : cd, &c. |
Therefore the polygons ABCDE, abcde are equiangular, and have the sides about their equal angles proportional. Hence they are similar.