First. Since the planes ABC, abc, are parallel, their intersections AB, ab, by the third plane SAB, are also parallel (B. VI., P. 10); hence, the triangles SAB, Sab, are similar (B. IV., P. 21), and we have

SA:

for a like reason, we have

A

B

SB: Sb :: SC: Sc;

and so on. Hence, the edges SA, SB, SC,

&c., are cut proportionally in a, b, c, &c.

The altitude SO is likewise cut in the same proportion, at the point o; for BO and bo are parallel, therefore, we have SO: So: SB: Sb.

Secondly. Since ab is parallel to AB, bc to BC, cd to CD, &c., the angle abc is equal to ABC, the angle bcd to BCD, and so on (B. VI., P. 13). Also, by reason of the similar triangles SAB, Sab, we have

AB : ab :: SB: Sb;

and by reason of the similar triangles SBC, Sbc, we have

hence,

SB : Sb :: BC : bc;

AB : ab :: BC : bc;

we might likewise have

BC : bc :: CD : cd,

and so on. Hence, the polygons ABCDE, abcde have their angles equal, each to each, and their sides, taken in the same order, proportional; hence, they are similar (B. IV., D. 1). Cor. 1. Let S-ABCDE, S-XYZ, be two pyramids, having a common vertex and their bases in the same plane; if these pyramids are cut by a plane parallel to the plane of their bases, the sections, abcde, xyz, will be to each other as the bases ABCDE, XYZ.

A

E

D

Χ

B

Y

For, the polygons ABCDE, abcde, being similar, their surfaces are as the squares of the homologous sides AB, ab; that is, B. IV., P. 27),

For the same reason,

-2

SA2:

2

XYZ : xyz: SX2: Soc2.

But since abc and xyz are in one plane, we have likewise (B. VI., P. 15),

hence,

SA : Sa :: SX: Soc;

ABCDE : abcde :: XYZ : xyz;

therefore, the sections abcde, xyz, are to each other as the bases ABCDE, XYZ.

Cor. 2. If the bases ABCDE, XYZ, are equivalent, any sections abcde, xyz, made at equal distances from the bases, are also equivalent.

PROPOSITION IV. THEOREM.

The convex surface of a right pyramid is equal to the perimeter of its base multiplied by half the slant height.

Let S be the vertex, ABCDE the base, and SF the slant height of a right pyramid; then will the convex surface be equal to SF×(AB+BC+CD+DE).

For, since the pyramid is right, the point O, in which the axis meets the base, is the centre of the polygon ABODE (D. 11); hence, the lines OA, OB, 00, &c., drawn to the vertices of the base, are equal.

E

e

In the right-angled triangles SAO, SBO, the bases and perpendiculars are equal: hence, the hypothenuses are equal: and it may be proved in the

same way, that all the edges of the right pyramid are

equal. The triangles, therefore, which form the convex surface of the prism are all equal to each other. But the area of either of these triangles, as ESA,

is equal to its base EA, multiplied by half the perpendicular SF, which is the slant height of the pyramid: hence, the area of all the triangles, or the convex surface of the pyramid, is equal to the perimeter of the base multiplied by half the slant height.

Cor. The convex surface of the frustum of a right pyramid is equal to half the sum of the perimeters of its upper and lower bases multiplied by its slant height.

For, since the section abcde is similar to the base (P. 3), and since the base ABCDE is a regular polygon (D. 11), it follows that the sides ea, ab, bc, cd, and de, are all equal to each other. Hence, the convex surface of the frustum ABCDE-d is composed of the equal trapezoids EAae, ABba, &c., and the perpendicular distance between the parallel sides of either of these trapezoids is equal to Ff, the slant height of the frustum. But the area of either of the trapezoids, as AEea, is equal to (EA + eα) × Fƒ (B. IV., P. 7): hence, the area of all of them, or the convex surface of the frustum, is equal to half the sum of the perimeters of the upper and lower bases multiplied by the slant height.

PROPOSITION V. THEOREM.

If the three faces which include a triedral angle of a prism are equal to the three faces which include a triedral angle of a second prism, each to each, and are like placed, the two prisms are equal.

Let B and b be the vertices of two triedral angles included by faces respectively equal to each other, and similarly placed; then will the prism ABCDE-K be equal to the prism abcde-k.

For, place the base abcde upon the equal base ABCDE;

then, since the triedral angles at b and B are equal, the parallelogram bh will coincide with BH, and the parallelogram bf with BF. But the two upper bases being equal to their corresponding lower bases, are equal to each other, and consequently, will coincide: hence, hi will coincide with HI, ik with IK, kf with KF; and therefore, the lateral faces of the prisms will coincide: hence, the two prisms coinciding throughout, are equal (a. 14).

Cor. Two right prisms, which have equal bases and equal altitudes, are equal. For, since the side AB is equal to ab, and the altitude BG to bg, the rectangle ABGF is equal to abgf; so also, the rectangle BGHỢ is equal to bghc; and thus the three planes, which include the triedral angle B, are equal to the three which include the triedral angle b. Hence, the two prisms are equal.

PROPOSITION VI. THEOREM.

In every parallelopipedon, the opposite faces are equal and

parallel.

Let ABCD be a parallelopipedon, then will its opposite

faces be equal and parallel.

For, the bases ABCD, EFGH, are equal parallelograms, and have their planes parallel (D. 7). It remains only

to show, that the same is true of any two opposite lateral faces, such as BCGF, ADHE.

A

E

H

F

G

B

Now, BC is equal and parallel to AD, because the base ABCD is a parallelogram; and since the lateral faces are also parallelograms, BF is equal and parallel to AE, and the same may be shown for the sides FG and EH, CG and DH; hence, the angle CBF is equal to the angle DAE, and the planes DAE, CBF, are parallel (B. VI., P. 13); and the parallelogram BCGF, is equal to the parallelogram ADHE. In the same way, it may be shown that the opposite parallelograms ABFE, DCGH, are equal and parallel.

Cor. 1. Since the parallelopipedon is a solid bounded by six faces, of which any two lying opposite to each other, are equal and parallel, it follows that any face and the one opposite to it, may be assumed as the bases of the parallelopipedon.

F

H

Cor. 2. The diagonals of a parallelopipedon bisect each other. For, suppose two diagonals BH, DF, to be drawn through opposite vertices. Draw also BD, FH. Then, since BF is equal and parallel to DH, the figure BDHF is a parallelogram; hence, the diagonals BH, DF, mutually bisect each other at E (B. I., P. 31). In the same manner, it may be shown that the diagonal BH and any other diagonal bisect each other; hence, the four diagonals mutually bisect each other, in a common point. If the six faces are equal to each other, this point may be regarded as the centre of the parallelopipedon.

B

E

D

Scholium. If three straight lines AB, AE, AD, passing through the same point A, and making given angles with each other, are known, a parallelopipedon may be formed on these lines. For this purpose, conceive a plane to be passed through the extremity of each line, and parallel to the plane of the other two, that is, through the point B pass a plane parallel to DAE, through D a plane parallel to BAE, and through E a plane parallel to BAD. The mutual intersections of these planes will form the parallelopipedon required.