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their homologous sides; that is, as AB' to ab'. Therefore, we have

Solid FD: solid fd:: AB'x AF: ab' xaf.

But since BF and bf are similar figures, their homologous sides are proportional; that is,

:

AB ab

AF: af,

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whence (Prop. X., B. II.),

Also

AF af AF af.

Therefore (Prop. XI., B. II.),

ABX AF: ab xaf:: AF: af:: AB': ab'.

Hence (Prop. IV., B. II.), we have

Solid FD: solid fd:: AB3: ab3 : : AF3: af3.

Therefore, similar prisms, &c.

PROPOSITION XIIL THEOREM.

If a pyramid be cut by a plane parallel to its base, 1st. The edges and the altitude will be divided proportionally. 2d. The section will be a polygon similar to the base.

Let A-BCDEF be a pyramid cut by a plane bcdef parallel to its base, and let AH be its altitude; then will the edges AB, AC, AD, &c., with the altitude AH, be divided proportionally in b, c, d, e, f, h; and the section bcdef will be similar to BCDEF.

First. Since the planes FBC, fbc are parallel, their sections FB, fb with a third F plane AFB are parallel (Prop. XII., B. VII.); therefore the triangles AFB, Afb are similar, and we have the proportion AF Af:: AB: Ab.

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For the same reason,

AB: Ab: AC: Ac,

B

D

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.),
FB: fb:: BC: bc.

For the same reason,

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.

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 Jases 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,

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

Therefore,

BCDEF: bcdef : : AB2 : Ab3.

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. XVI., B. VII.);

consequently, BCDEF: bcdef :: MNO : mno.

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

PRC POSITION XIV. THEOREM.

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.

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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 distances 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.

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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 pyramid. In the same manner, upon the triangles EFG, HİK, &c., taken as bases, construct exterior prisms, having for edges the parts EH, HL, &c., of the line AB. In like man

ner, 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 prisma

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.

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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 para.lelogram ACFD. Draw the diagonal CD, and through the points C, D, E pass a plane, dividing the quadrangular pyramid into two triangular ones E-ACD, E-CFD. Then, because ACFD is a parallelogram, of which

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