Divide the altitude AT into equal parts Ax, xy, &c., each of which is less than Aa, and let k denote one of these parts; through the points of division pass planes parallel to the plane of the bases; the sections of the two pyramids, by each of these planes, will be equal, namely, DEF to def, GHI to ghi, &c. (P. III., C. 2).

On the triangles ABC, DEF, &c., taken as lower bases, construct exterior prisms whose edges shall be parallel to AS, and whose altitudes shall be equal to : and on the triangles def, ghi, &c., taken as upper bases, construct interior prisms, whose edges shall be parallel to Sa, and whose altitudes shall be equal to k. It is evident that the sum of the exterior prisms is greater than the pyramid S-ABC, and also that the sum of the interior prisms is less than the pyramid S-abc: hence, the difference between the sum of the exterior and the sum of the interior prisms, is greater than the difference between the two pyramids.

Now, beginning at the bases, the second exterior prism EFD-G, is equal to the first interior prism efd-a,

because they have the same altitude k, and their bases EFD, efd, are equal: for a like reason, the third exterior prism HIG-K, and the second interior prism hig-d, are equal, and so on to the last in each set: hence, each of the exterior prisms, excepting the first BCA-D, has an equal corresponding interior prism; the prism BCA-D, is, therefore, the difference between the sum of all the exterior prisms, and the sum of all the interior prisms. But the difference between these two sets of prisms is greater than that between the two pyramids, which latter difference was supposed to be equal to a prism whose base is BCA, and whose altitude is equal to Aa, greater than k; consequently, the prism BCA-D is greater than a prism having the same base and a greater altitude, which is impossible : hence, the supposed inequality between the two pyramids cannot exist; they are, therefore, equal in volume; which was to be proved.

Any triangular prism may be divided into three triangular pyramids, equal to each other in volume.

altitude, because their bases are in the same plane AD, and their vertices at the same point F; hence, they are equal in volume (P. XV.). The pyramids ABC-F and DEF-C, have their bases ABC and DEF, equal, because they are the bases of the given prism, and their altitudes are equal because each is equal to the altitude of the prism; they are, therefore, equal in volume: hence, the three pyramids into which the prism is divided, are all equal in volume ; which was to be proved.

Cor. 1. A triangular pyramid is one-third of a prism, having an equal base and an equal altitude.

Cor. 2. The volume of a triangular pyramid is equal to one-third of the product of its base and altitude.

PROPOSITION XVII. THEOREM.

The volume of any pyramid is equal to one-third of the product of its base and altitude.

Let S-ABCDE, be any pyramid: then is its volume equal to one-third of the product of its base and altitude. For, through any lateral edge, as SE,

pass the planes SEB, SEC, dividing the pyramid into triangular pyramids. The altitudes of these pyramids will be equal to each other, because each is equal to that of the given pyramid. Now, the volume of each triangular pyramid is equal to onethird of the product of its base and altitude (P. XVI., C. 2); hence, the sum of the volumes of the triangular pyramids, is

A

E

B

equal to one-third of the product of the sum of their bases

by their common altitude.

But the sum of the triangular pyramids is equal to the given pyramid, and the sum of their bases is equal to the base of the given pyramid: hence, the volume of the given pyramid is equal to onethird of the product of its base and altitude; which was to be proved.

Cor. 1. The volume of a pyramid is equal to one-third of the volume of a prism having an equal base and an equal altitude.

Cor. 2. Any two pyramids are to each other as the products of their bases and altitudes. Pyramids having equal bases are to each other as their altitudes. Pyramids having equal altitudes are to each other as their bases.

Scholium. The volume of a polyedron may be found by dividing it into triangular pyramids, and computing their volumes separately. The sum of these volumes will be equal to the volume of the polyedron.

PROPOSITION XVIII. THEOREM.

The volume of a frustum of any triangular pyramid is equal to the sum of the volumes of three pyramids whose common altitude is that of the frustum, and whose bases are the lower base of the frustum, the upper base of the frustum, and a mean proportional between the two bases.

Let FGH-h be a frustum of any triangular pyramid : then will its volume be equal to that of three pyramids whose common altitude is that of the frustum, and whose bases are the lower base FGH, the upper base fgh, and a mean proportional between their bases.

For, through the edge FH, pass the plane FHg, and through the edge fg, pass the plane fgH, dividing the frustum into three pyramids. The pyramid g-FGH, has for its base the lower base FGH of the frustum, and its altitude is equal to that of the frustum, because its vertex g, is in the plane of the upper base. The pyramid H-fgh, has for its base the upper base fgh of the frustum, and its altitude is equal to that of the frustum, because its vertex lies in the plane of the lower base.

F

K

G

H

The remaining pyramid may be regarded as having the triangle FfH for its base, and the point g for its vertex. From g, draw gK parallel to fF, and draw also KH and Kf. Then will the pyramids K-FfH and g-FfH, be equal; for they have a common base, and their altitudes are equal, because their vertices K and g are in a line parallel to the base (B. VI., P. XII., C. 2).

Now, the pyramid K-FfII may be regarded as having FKH for its base and ƒ for its vertex. From K, draw KL parallel to GH; it will be parallel to gh: then will the triangle FKL be equal to fgh, for the side FK is equal to fg, the angle F to the angle f, and the angle K to the angle g. But, FKH is a mean proportional between FKL and FGHI (B. IV., P. XXIV., C.), or between fgh and FGH. The pyramid f-FKH, has, therefore, for its base a mean proportional between the upper and lower bases of the frustum, and its altitude is equal to that of the frus tum; but the pyramid f-FKH is equal in volume to the pyramid g-FFH: hence, the volume of the given frustum is equal to that of three pyramids whose common altitude is equal to that of the frustum, and whose bases are the upper base, the lower base, and a mean proportional between them; which was to be proved.