And conceive AB to be divided into m equal parts, or rectangles, AI, LK, MB (by dividing AN into that number of equal parts, and drawing IL. KM, parallel to BN). And let EF be divided, in like manner, into n equal parts, or rectangles, EO, PF: all of these parts of both bases being mutually equal among themselves. And through the lines of division let the plane sections LR, MS, PV, pass parallel to AQ, ET.

Then the parallelopipedons AR, LS, MC, EV, PG, are all equal, having equal bases and heights. Therefore, the solid AC is to the solid EG, as the number of parts in AC to the number of equal parts in EG, or as the number of parts in AB to the number of equal parts in EF; that is, as the base AB to the base EF. Q. E. D.

Corol. From this proposition, and the corollary to the last, it appears, that all prisms and cylinders of equal altitudes, are to each other as their bases; every prism and cylinder being equal to a rectangular parallelopipedon of an equal base and height.

PROP. V.

Rectangular parallelopipedons, of equal bases, are in proportion to each other as their altitudes.

Let AB, CD, be two rectangular parallelopipedons standing on the equal bases AE, CF; then will AB be to CD as the altitude EB is to the altitude DF.

For, let AG be a rectangular paralielopipedon on the base AE, and its altitude EG equal to the altitude FD of the solid CD.

Then, AG and CD are equal, being prisms

B

E

H

C

of equal bases and altitudes. But if HB, HG, be considered as bases, the solids AB, AG, of equal altitude AH, will be to each other as those bases HB, HG. But these bases HB, HG, being parallelograms of equal altitude HE, are to each other as their bases EB, EG; and therefore the two prisms AB, AG, are to each other as the lines EB, EG. But AG is equal CD, and EG equal FD; consequently, the prisms AB, CD, are to each other as their altitudes EB, FD; that is, AB: CD :: EB: FD. Q. E. D.

Corol. 1. From this proposition, and the corollary to Prop. III., it appears, that all prisms and cylinders, of equal bases, are to one another as their altitudes. Corol. 2. Because, by corol. 1, prisms and cylinders are as their altitudes, when their bases are equal. And, by the corollary to the last theorem, they are as their bases, when their altitudes are equal. Therefore, universally, when neither are equal, they are to one another as the product of their bases and altitudes. And hence, also, these products are the proper numeral measures of their quantities or magnitudes.

PROP. VI.

Similar prisms and cylinders are to each other as the cubes of their altitudes, or of any other like linear dimensions.

Let ABCD, EFGH, be two similar prisms; then will the prism CD be to the prism GH, as AB to EF, or as AD3 to EH3.

For, the solids are to each other as the product of their bases and altitudes (Prop. v., cor. 2), that is, as AC. AD to EG. EH. But the bases, being

H

similar planes, are to each other as the squares of their like sides, that is, AC to EG as AB to EF; therefore, the solid CD is to the solid GH as AB. AD to EF". EH. But BD and FH, being similar planes, have their like sides proportional, that is, AB: EF :: AD: EH, or AB2: EF2 :: AD2: EH2; therefore, AB. AD: EF'. EH :: AB3: EF3, or :: AD3: EH3; and consequently, the solid CD: solid GH :: AB': EF3 :: AD3: EH3 Q. E. D.

PROP. VII

In a pyramid, a section parallel to the base is similar to the base, and these two planes will be to each other as the squares of their distances from the vertex. Let ABCD be a pyramid, and EFG a section parallel to

the base BCD, also AIH a line perpendicular to the two

planes at H and I; planes, and the plane to AI.

then will BD, EG, be two similar

BD will be to the plane EG as AH2

G

F

D

H

For, join CH, FI. Then, because a plane cutting two parallel planes, makes parallel sections, therefore the plane ABC, meeting the two parallel planes BD, EG, makes the sections BC, EF, parallel;—in like manner, the plane ACD makes the sections CD, FG, parallel. Again, because two pair of parallel lines make equal angles, the two EF, FG, which are parallel to BC, CD, make the angle EFG equal the angle BCD. And, in like manner, it is shown, that each angle in the plane EG is equal to each angle in the plane BD, and consequently those two planes are equiangular.

Again, the three lines AB, AC, AD, making with the parallels BC, EF, and CD, FG, equal angles; and the angles at A being common, the two triangles ABC, AEF, are equiangular, as also the two triangles ACD, AFG and have therefore their like sides proportional, namely, AC:AF:: BC: EF.: CD : FG. And, in like manner, it may be shown, that all the lines in the plane EG are proportional to all the corresponding ones in the base BD. Hence, these two planes, having their angles equal and their sides proportional, are similar.

But, similar planes being to each other as the squares of their like sides, the plane BD EG :. BC: EF": or :: AC: AF, by what is shown above. But the two triangles AHC, AIF, having the angles H and I right ones, and the angle A common, are equiangular, and have therefore their like sides proportional, namely, AC: AF:: AH: AI, or AC : AF* :: AH2; AI2. Consequently the two planes BD, EG, which are as the former squares AC, AF3, will be also as the latter squares AH, AI, that is, BD: EG:: AH2: AI2.

PROP. VIII.

In a right cone a section parallel to the base is a circle; and this section is to the base as the squares of their distances from the vertex.

Let ABCD be a right cone, and GHI a section parallel to the base BCD; then will GHI be a circle, and BCD, GHI, will be to each other as the squares of their distances from the vertex.

For, draw ALF perpendicular to the two parallel planes; and let the planes ACE, ADE, pass through the axis of the cone AKE, meeting the section in the three points H, I, K.

B

Then, since the section GHI is parallel to the base BCD, and the planes CK, DK, meet them, HK is parallel to CE, and IK to DE. And because the triangles formed by these lines are equianglar, KH : EC :: AK : AE :: KI : ED. But EC is equal to ED, being radii of the same circle; therefore, KI is also equal to KH. And the same may be shown of any other lines drawn from the point K to the circumference of the section GHI, which is therefore a circle. Again, by similar triangles, AL : AF :: AK: AE or :: KI: ED, hence AL': AF :: KI: ED; but, KI: ED':: circle GHI: circle BCD; therefore, AL: AF:: circle GHI: circle BCD. Q. E. D.

PROP. IX.

If a right cone BCD be cut by a plane AGK which is parallel to a plane touching the cone along the slant side BC, the section AGK is a parabola.

Let BCD be that position of the generating triangle, which is perpendicular to the cutting plane AGK; AH their common section, which is parallel to BC. Draw AL parallel to CD. Then, since the plane BCD passes through the axis, it is perpendicular to the base CKD and to every circular section EPF parallel to the base; it is also perpendicular to AGK. Hence, the common section PR of the planes AGK, EPF, is perpendicular to BCD and therefore to AH and EF.

But, AN: NF :: BC: CD, which is a constant ratio, therefore AN & NF

stant) & NP by the

whose axis is AH.

R

E

G

H

EN × NF (for EN is equal and parallel to AL, and conproperty of the circle. Hence, the curve is a parabola, (See Conic Sections, infra, Parabola, Prop. VII.)

Cor. If L be the latus rectum of the parabola GAK, L × AN = NP' = EN X NF.

If a right cone BAD be cut by a plane AMP through both slant sides, the section is an ellipse.

Let BAD be that position of the generating triangle which is perpendicular to the cutting plane; EPF any circular section. Draw MHK parallel to AD and therefore bisected by the axis BO.

Then, AN EN :: AM: MK

:

NM NF :: AM: AD;

.. AN × NM: EN × NF (NP2) :: AM2: AD × MK

which is the property of an ellipse, one of whose axes is AM and the other a mean proportional between AD and MK. (Conic Sections, infra, Ellipse, Prop. XII.)

M

PROP. XI.

If a right cone BED be cut through one side BE by a plane RAP which being produced backwards cuts the other side DB produced, the section is an hyperbola.

Let DGEH be any circular section, BGH a triangular section through the vertex B of the cone parallel to the plane RAP.

Then, AN EN :: BF: EF

NM: ND :: BF: FD

AN X NM: EN × ND (NP2) :: BF* : EF × FD (FH3) which is the property of an hyperbola, whose axis major is AM and whose conjugate axis is to AM as FH to BF.

Cor. If GT, HT, be tangents to the circle at G, H; and planes passing through GT, HT, respectively, touch the cone along the lines BG, BH; also if TB, the common section of the planes, meet AM in C: then the common sections CO, CQ, of the plane RAP extended to meet the tangent planes are the asymptotes of the hyperbola.

Draw BL parallel to DE, meeting AM in L. Then the axes of the hyperbola being in the proportion of BF to FH, the angle GBH or the equal angle OCQ is the angle between the asymptotes.

Now, by similar triangles ALB, BFE, and CLB, BFT; AL: CL :: TF : FE, and therefore AC: CL:: TE: FE. In like manner, by similar triangles MLB, BFD, and CLB, BFT; ML: CL:: TF: DF, and therefore CM: CL :: TD: DF. But by the property of the circle, TE : FE :: TD : DF. Therefore, CA CM. Hence C is the centre of the hyperbola, and CO, CQ, are the asymptotes. (Conic Sections, infra, Hyperbola, Prop. XII.)

PROP. XIT.

All pyramids and right cones of equal bases and altitudes are equal to one another. Let ABC, DEF, be any pyra

mids and cone, of equal bases BC, EF, and equal altitudes AG, DH; then will the pyramids and cone ABC and DEF, be equal.

For, parallel to the bases, and at equal distances AN, DO, from the vertices, suppose the planes IK, LM, to be drawn.

AAA

Then, by Props. vII. and VIII.,

DO2 : DH2 :: LM : EF,

and AN AG :: IK : BC.

But, since ANꞌ, AG2, are equal to DO3, DH'; therefore, IK : BC :: LM : EF But BC is equal to EF, by hypothesis; therefore, IK is also equal to LM.

In the same manner, it is shown that any other sections, at equal distance from the vertex, are equal to each other.

Since, then, every section in the cone, is equal to the corresponding section in the pyramids, and the heights are equal, the solids ABC, DEF, composed of those sections, must be equal also. Q. E. D.

PROP. XIII.

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

Let ABCDEF be a prism, and BDEF a pyramid, upon the same triangular base DEF; then will the pyramid BDEF be a third part of the prism ABCDEF.

For, in the planes of the three sides of the prism, draw the diagonals BF, BD, CD. Then the two planes BDF, BCD, divide the whole prism into the three pyramids BDEF, DABC, DBCF; which are proved to be all equal to one another as follows:

Since the opposite ends of the prism are equal to each other, the pyramid whose base is ABC and vertex D, is equal to the pyramid whose base is DEF and vertex B (Prop. XII.), being pyramids of equal base and altitude.

But the latter pyramid, whose base is DEF and vertex B, is the same solid as the pyramid whose base is BEF and vertex D, and this is equal to the third pyramid, whose base is BCF and vertex D, being pyramids of the same altitude and equal bases BEF, BCF.

Consequently, all the three pyramids which compose the prism, are equal to each other, and each pyramid is the third part of the prism, or the prism is triple of the pyramid. Q. E. D.

Corol. 1. Every pyramid, whatever its figure may be, is the third part of a prism of the same base and altitude; since the base of the prism, whatever be its figure, may be divided into triangles, and the whole solid into triangular prisms and pyramids.

Cor. 2. Any right cone is the third part of a cylinder, or of a prism, of equal base and altitude; since it has been proved that a cylinder is equal to a prism, and a cone equal to a pyramid, of equal base and altitude.

SCHOLIUM.—Whatever has been demonstrated of the proportionality of prisms, or cylinders, holds equally true of pyramids or cones, the former being always triple the latter; viz. that similar pyramids or cones, are as the cubes of their like linear sides, or diameters, or altitudes, &c.