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 regular pyramid is equal to half the perimeters of its upper and lower bases multiplied by its slant height.

For, since the section abcde is similar to the base (Prop. III.), and since the base ABCDE is a regular polygon (Def. 14.), 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 formed by 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+ea) x Ff (Book IV. Prop. VII.): hence the area of all of them, or the convex surface of the frustum, is equal to half the perimeters of the upper and lower bases multiplied by the slant height.

PROPOSITION V. THEOREM.

If the three planes which form a solid angle of a prism, are equal to the three planes which form the solid angle of another prism, each to each, and are like situated, the two prisms will be equal to each other.

Let the base ABCDE be equal to the base abcde, the parallelogram ABGF equal to the parallelogram abgf, and the parallelogram BCHG equal to bchg; then will the prism ABCDE-K be equal to the prism abcde-k.

For, lay the base ABCDE upon its equal abcde; these two bases will coincide. But the three plane angles which form

the solid angle B, are respectively equal to the three plane angles, which form the solid angle b, namely, ABC=abc, ABG abg, and GBC=gbc; they are also similarly situated. hence the solid angles B and bare equal (Book VI. Prop. XXI, Sch.); and therefore the side BG will fall on its equal bg. It is likewise evident, that by reason of the equal parallelograms ABGF, abgf, the side GF will fall on its equal gf, and in the same manner GH on gh; hence, the plane of the upper base, FGHIK will coincide with the plane fghik (Book VI. Prop. II.).

But the two upper bases being equal to their corresponding lower bases, are equal to each other: hence HI will coincide with hi, IK with ik, and KF with kf; and therefore the lateral faces of the prisms will coincide: therefore, the two prisms coinciding throughout are equal (Ax. 13.).

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 will be equal to abgf; so also will the rectangle BGHC be equal to bghc; and thus the three planes, which form the solid angle B, will be equal to the three which form the solid angle b. Hence the two prisms are equal.

PROPOSITION VI. THEOREM.

In every parallelopipedon the opposite planes are equal and parallel.

By the definition of this solid, the bases ABCD, EFGH, are equal parallelograms, and their sides are parallel: it remains only to show, that the same is true of any two opposite lateral faces, such as AEHD, BFGC. Now AD is equal and parallel to BC, because the figure ABCD is a par

allelogram; for a like reason, AE is parallel to BF: hence the angle DAE is equal to the angle CBF, and the planes DAE, CBF, are parallel (Book VI. Prop. XIII.); hence also the parallelogram DAEH is equal to the parallelogram CBFG. In the same way, it might be shown that the opposite parallelograms ABFE, DCGH, are equal and parallel.

Cor. 1. Since the parallelopipedon is a solid bounded by six planes, whereof those 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.

Cor. 2. The diagonals of a parallelopipedon bisect each other. For, suppose two diagonals EC, AG, to be drawn both through opposite vertices: since AE is equal and parallel to CG, the figure AEGC is a parallelogram; hence the diagonals EC, AG will mutually bisect each other. In the same manner, we could show that the diagonal EC and another DF bisect each other; hence the four diagonals will mutually bisect each other, in a point which may be regarded as the centre of the parallelopipedon.

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 those lines. For this purpose, a plane must be passed through the extremity of each line, and parallel to the plane of the other two; that is, through the point_B 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.

PROPOSITION VII. THEOREM.

The two triangular prisms into which a parallelopipedon is divided by a plane passing through its opposite diagonal edges, are equivalent.

Let the parallelopipedon ABCD-H beat mela divided by the plane BDHF passing through its diagonal edges: then will the triangular prism ABD-H be equivalent to the triangular prism BCD-H.

Through the vertices B and F, draw the planes Badc, Fehg, at right angles to the side BF, the former meeting AE, DH, CG, A the three other sides of the parallelopipedon, in the points a, d, c, the latter in e, h, g: the sections Badc, Fehg, will be equal parallelograms. They are equal, because

a

E

B

G

they are formed by planes perpendicular to the same straight line, and consequently parallel (Prop. II.); they are parallelograms, because aB, dc, two opposite sides of the same section, are formed by the meeting of one plane with two parallel planes ABFE, DCGH.

For a like reason, the figure BaeF is a parallelogram; so also are BFgc, cdhg, adhe, the other lateral faces of the solid Badc-g; hence that solid is a prism (Def. 6.); and that prism is right, because the side BF is perpendicular to its base.

But the right prism Badc-g is divided by the plane BH into two equal right prisms Bad-h, Bcd-h; for, the bases Bad, Bcd, of these prisms are equal, being halves of the same parallelogram, and they have the common altitude BF, hence they are equal (Prop. V. Cor.).

It is now to be proved that the oblique triangular prism ABD-H will be equivalent to the right triangular prism Bad-h; and since those prisms have a common part ABD-h, it will only be necessary to prove that the remaining parts, namely, the solids BaADd, FeEHh, are equivalent.

Now, by reason of the parallelograms ABFE, aBFe, the sides AE, ae, being equal to their parallel BF, are equal to each other; and taking away the common part Ae, there remains Aa Ee. In the same manner we could prove Dd=Hh.

Next, to bring about the superposition of the two solids BaADd, FeEHh, let us place the base Feh on its equal Bad: the point e falling on a, and the point h on d, the sides eE, H, will fall on their equals aA, dD, because they are perpendicu lar to the same plane Bad. Hence the two solids in question will coincide exactly with each other; hence the oblique prism BAD-H, is equivalent to the right one Bad-h.

In the same manner might the oblique prism BCD-H, be proved equivalent to the right prism Bcd-h. But the two right prisms Bad-h, Bcd-h, are equal, since they have the same altitude BF, and since their bases Bad, Bdc, are halves of the same parallelogram (Prop. V. Cor.). Hence the two trian

gular prisms BAD-H, BDC-G, being equivalent to the equal right prisms, are equivalent to each other.

Cor. Every triangular prism ABD-HEF is half of the parallelopipedon AG described with the same solid angle A, and the same edges AB, AD, AE.

PROPOSITION VIII. THEOREM.

If two parallelopipedons have a common base, and their upper bases in the same plane and between the same parallels, they will be equivalent.

Let the parallelopipedons AG, AL, have the common base AC, and their upper bases EG, MK, in the same plane, and between the same parallels HL, EK; then will they be equivalent.

There may be three cases, according as EI is

greater, less than, or equal to, EF; but the demonstration is the same for all. In the first place, then we shall show that the triangular prism AEI-MDH, is equal to the triangular prism BFK-LCG.

Since AE is parallel to BF, and HE to GF, the angle AEI =BFK, HEI=GFK, and HEA-GFB. Also, since EF and IK are each equal to AB, they are equal to each other. To each add FI, and there will result EI equal to FK: hence the triangle AEI is equal to the triangle BFK (Bk. I. Prop. V), and the paralellogram EM to the parallelogram FL. But the parallelogram AH is equal to the parallelogram CF (Prop. VI): hence, the three planes which form the solid angle at E are respectively equal to the three which form the solid angle at F, and being like placed, the triangular prism AEI-M is equal to the triangular prism BFK-L.

But if the prism AEI-M is taken away from the solid AL, there will remain the parallelopipedon BADC-L, and if the prism BFK-L is taken away from the same solid, there will remain the parallelopipedon BADC-G; hence those two paral lelopipedons BADC-L, BADC-G, are equivalent.