and so for the other edges. Therefore the edges AB, AC, &c., are cut proportionally in b, c, &c. Also, since BH and bh are parallel, we have

AH: Ah: AB : Ab.

Secondly Because fb is parallel to FB, bc to BC, cdo CD, &c., the angle fbc is equal to FBC (Prop. XV., B. VII.), the angle bed is equal to BCD, and so on. Moreover, since the triangles AFB, Afb are similar, we have

FB: fb:: AB· Ab.

And because the triangles ABC, Abc are similar, we have AB: Ab:: BC: bc.

Therefore, by equality of ratios (Prop. IV., B. II.),

For the same reason,

FB: fb BC: bc.

BC: bc:: CD : cd, and so on.

Therefore the polygons BCDEF, bcdef have their angles equal, each to each, and their homologous sides proportional, hence they are similar. "Therefore, if a pyramid, &c.

A

Cor. 1. If two pyramids, having the same altitude, and their bases situated in the same plane, are cut by a plane parallel to their bases, the sections will be to each other as the bases. Let A-BCDEY, A-MNO

be two pyramids having the same altitude, and their ases situated in the same plane; if these pyramids are cut by a plane parallel to the bases, the sections bcdef, mno will be to each other as the bases BCDEF, MNO.

For, since the polygons BCDEF, bcdef are similar,

their surfaces are as the squares of the homologous sides BC hc (Prop. XXVI., B. IV.). But, by the preceding Proposition BC: bc:: AB: Ab.

Therefore,

BCDEF: bcdef : : AB2 : Ab2.

For the same reason,

MNO: mno :: AM2 : Am2.

But since bcdef and mno are in the same plane, we have
AB: Ab: AM : Am (Prop. XVİ., B. VII.);

consequently, BCDEF: bcdef:: MNO: mno.

Cor. 2. If the bases BCDEF, MNO are equivalent, the sections bcdef, mno will also be equivalent,

PROPOSITION XIV. THECREM.

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

Let A-BDE be a regular pyramid, whose base is the polygon BCDEF, and its slant height AH; then will its convex surface be equal to the perimeter BC+CD+DE, &c., multiplied by half of AH.

D

The triangles AFB, ABC, ACD, &c., are all equal for the sides FB, BC, CD, &c., are all equal, (Def. 13); and since the oblique, lines AF, AB, AC, &c., are all at equal dis-" tances from the perpendicular, they are H equal to each other (Prop. V., B. VII.). Hence the altitudes of these several triangles are equal. But the area of the triangle AFB is equal to FB, multiplied by half of AH; and the same is true of the other triangles ABC, ACD, &c. Hence the sum of the triangles is equal to the sum of the bases FB, BC, CD, DE, EF, multiplied by half the common altitude AH; that is, the convex surface of the pyramid is equal to the perimeter of its base, multiplied by half the slant height.

Cor. 1. The convex surface of a frustum of a regular pyramid is equal to the sum of the perimeters of its two bases, multiplied by half its slant height.

Each side of a frustum of a regular pyramid, as FBbf, is a trapezoid (Prop. XIII.). Now the area of this trapezoid is equal to the sum of its parallel sides FB, fb, multiplied by half its altitude Hh (Prop. VII., B. IV.). But the altitude of each of these trapezoids is the same; therefore the area of all the trapezoids, or the convex surface of the frustum, is equal to the sum of the perimeters of the two bases, multiplied by half the slant height.

Cor. 2. If the frustum is cut by a plane, parallel to the bases, and at equal distances from them, this plane must bisect the edges Bb, Cc, &c. (Prop. XVI., B. IV.); and the area of each trapezoid is equal to its altitude, multiplied by the line which joins the middle points of its two inclined sides (Prop. VII., Cor., B. IV.). Hence the convex surface of a frustum of a pyramid is equal to its slant height, multiplied by the perimeter of a section at equal distances between the two bases.

PROPOSITION XV. THEOREM.

Triangular pyramids, having equivalent bases and equal al rtudes, are equivalent.

Let A-BCD, a-bcd be two triangular pyramids having equivalent bases BCD, bcd, supposed to be situated in the same plane, and having the common altitude TB; then will the pyramid A-BCD be equivalent to the pyramid a-bcd.

For, if they are not equivalent, let the pyramid A-BCD exceed the pyramid a-bcd by a prism whose base is BCD, and altitude BX.

Divide the altitude BT into equal parts, each less than BX; and through the several points of division,let planes be made to pass parallel to the base BCD, making the sections EFG, efg equivalent to each other (Prop. XIII., Cor. 2): also, HİK equivalent to hik, &c.

From the point C, draw the straight line CR parallel to BE, meeting EF produced in R; and from D draw DS parallel to BE, meeting EG in S. Join RS, and it is plain that the solid BCD-ERS is a prism lying partly without the pyr amid. In the same manner, upon the triangles EFG, HIK, &c., taken as bases, construct exterior prisms, having for edges the parts EH, HL, &c., of the line AB. In like manner, on the bases efg, hik, lmn, &c., in the second pyramid, construct interior prisms, having for edges the corresponding parts of ab. It is plain that the sum of all the exterior prisms

of the pyramid A-BCD is greater than this pyramid; and. also, that the sum of all the interior prisms of the pyramid a-bcd is smaller than this pyramid. Hence the difference between the sum of all the exterior prisms, and the sum of all the interior ones, must be greater than the difference be tween the two pyramids themselves.

Now, beginning with the bases BCD, bcd, the second ex terior prism EFG-H is equivalent to the first interior prism efg-b, because their bases are equivalent, and they have the same altitude. For the same reason, the third exterior prism HIK-L and the second interior prism hik-e are equivalent; the fourth exterior and the third interior; and so on, to the last in each series. Hence all the exterior prisms of the pyramid A-BCD, excepting the first prism BCD-E, have equivlent corresponding ones in the interior prisms of the pyramid a-bcd. Therefore the prism BCD-E is the difference between the sum of all the exterior prisms of the pyramid A-BCD, and the sum of all the interior prisms of the pyramid a-bcd. But the difference between these two sets of prisms has been proved to be greater than that of the two pyramids; hence the prism BCD-E is greater than the prism BCD-X; which is impossible, for they have the same base BCD, and the altitude of the first, is less than BX, the altitude of the second. Hence the pyramids A-BCD, a-bcd are not unequal; that is, they are equivalent to each other. There fore, triangular pyramids, &c.

PROPOSITION XVI. THEOREM.

Every triangular pyramid is the third part of a triangular prism having the same base and the same altitude.

Let E-ABC be a triangular pyramid, and ABC-DEF a triangular prism having the same base and the same altitude; then will the pyramid be one third of the prism.

F

Cut off from the prism the pyramid E-ABC by the plane EAC; there will remain the solid E-ACFD, which may be considered as a quadrangular pyramid whose vertex is E, and whose base is the paralelogram ACFD. Draw the diagohal CD, and through the points C, D, E pass a plane, dividing he quadrangular pyramid into two triangular ones E-ACD E-CFD. Then, because ACFD is a parallelogram, of which

CD is the aiagonal, the triangle ACD is Therefore equal to the triangle CDF.

A

the pyramid, whose base is the triangle ACD, and vertex the point E, is equivalent to the pyramid whose base is the triangle CDF, and vertex the point E. But the latter pyramid is equivalent to the pyramid E-ABC for they have equal bases, viz., the triangles ABC, DEF, and the same altitude, viz., the altitude of the prism ABC-DEF. Therefore the three pyramids E-ABC, E-ACD, E-CDF, are equivalent to each other, and they compose the whole prism ABC-DEF; hence the pyramid E-ABC is the third part of the prism which has the same base and the same altitude.

Cor. The solidity of a triangular pyramid is measured by the product of its base by one third of its altitude.

PROPOSITION XVII. THEOREM.

The solidity of every pyramid is measured by the product of its base by one third of its altitude.

Let A-BCDEF be any pyramid, whose base is the polygon BCDEF, and altitude AH; then will the solidity of the pyramid be measured by BCDEFAH.

B

A

Divide the polygon BCDEF into triangles by the diagonals CF, DF; and let planes pass through these lines and the vertex A; they will divide the polygonal pyramid A-BCDEF into triangular pyramids, all having the same altitude AH. But each of these pyramids is measured by the product of its base by one third of its altitude (Prop. XVI., Cor.); hence the sum of the triangular pyramids, or the polygonal pyramid A-BCDEF, will be measured by the sun of the triangles BCF, CDF, DEF, or the polygon BCDEF, multiplied by one third of AH. Therefore every pyramid is measured by the product of its base by one third of its altitude.

Cor. 1. Every pyramid is one third of a prism having the same base and altitude.

Cor. 2. Pyramids of the same altitude are to each other as their bases; pyramids of the same base are to each other