348

AC AF BC: EF

:: CD: FG.

And in like manner it may be shown, that all the lines in the plane FG, are proportional to all the corresponding lines in-the base BD. Hence these two planes, having their angles equal, and their sides proportional, are similar, by def. 68.

But, similar planes being to each other as the squares of their like sides, the plane BD EG BC2: EF2, or AC2: Also, the two triangles AF2, by what is shown above.

AHC,

AIF. having the angles H and 1 right ones (th. 98), and the angle A common, are equiangular, and have therefore their like sides proportional, namely, AC: AF :: AH: AI, Consequently the two planes or AC2 AF2 :: AH? : A12. as the former squares Ac2, AF2, BD, EG, which are be also as the latter squares AH, AI, that is, BD EG: AH2: AI2.

will

Q. E. D.

THEOREM CXIIL

In a Cone, any 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 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.

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 (th 105). And because the triangles formed by these lines are equiangular, 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 perimeter of the section GHI, which is therefore a circle (def. 45).

Again, by similar triangles, AL: AF :: AK: AE or :: KI ED, hence AL2: AF2 :: KI2: ED2; but K12: ED2 :: circle GHI circle BCD (th. 93); therefore AL2: AF2 circle GHI circle BCD.

::

Q. E. D. THEOREM

THEOREM CXIV.

All Pyramids, and Cones, of Equal Bases and Altitudes, are Equal to one another.

For parallel to the

bases and at equal distances AN, DO, from the vertices, sup

pose the planes IK, LM, to be drawn.

Then, by the two preceding theorems,

D02: DH2: LM: EF, and

AN2 AG2 :: IK: BC.

But since AN, AG2, are equal to Do3, DH2, therefore IK: BC :: LM: EF.

But BC

is equal to EF,

by hypothesis; therefore IK is also equal to LM.

In like 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 all those sections, must be equal also.

Q. E. D.

THEOREM CXV.

Every Pyramid is the Third Part of a Prism of the Same Base and Altitude.

LET ABCDEF be a prism, and BDEF a pyramid, on the same triangular base DEF then will the pyramid, BDEF be a third part of the prism ABCDEF.

A

B

For, in the planes of the three sides of

the prism, draw the diagonals BF, BD, CD.

E

Then the two planes BDF, BCD, divide the

F

whole prism into the three pyramids BDEF, DAEC, 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.

pyramid whose base is DEF and vertex B (th. 114), being pyramids of equal base and altitude.

But the latter pyramid, whose base is DEF and vertex в, 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.

B

E

A

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.

Hence also, 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.

Corol. Any 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. And the same for all similar solids, whatever, viz. that they are in proportion to each other, as the cubes of their like linear dimensions, since they are composed of pyramids every way similar.

THEOREM CXVI.

If a Sphere be cut by a Plane, the Section will be a Circle.

Let the sphere AEBF be cut by the plane ADB; then will the section ADB be a circle.

Draw the chord AB, or diameter of the section; perpendicular to which, or to the section ADB, draw the axis of the sphere ECGF, through the centre c, which will bisect the chord AB in the point (th. 41). Also, join CA, CB;

E

F

and

and draw CD, GD, to any point D in the perimeter of the section ADB.

Then, because co is perpendicular to the plane ADE, it is perpendicular both to GA and GD (def. 90). So that CGA, CGD, are two right-angled triangles, having the perpendicular co common, and the two hypothenuses CA, CD, equal, being both radii of the sphere; therefore the third sides GA, GD, are also equal (cor. 2, th. 34). In like manner it is shown, that any other line, drawn from the centre & to the circumference of the section ADB, is equal to GA or GB; consequently that section is a circle.

Corol. The section through the centre, is a circle having the same centre and diameter as the sphere, and is called a great circle of the sphere; the other plane sections being little circles.

THEOREM CXVII.

Every Sphere is Two-Thirds of its Circumscribing Cylinder

LET ABCD be a cylinder, circumscribing the sphere EFGH; then will the sphere EFGH be two-thirds of the cylinder ABCD.

E

D

K

G

For, let the plane ac be a section of the sphere and cylinder through the centre 1. Join AI, BI.. Also, let FIH be parallel to AD or BC, and EIG and KL parallel to AB or DC, the base of the cylinder; the latter line KL meeting BI in м, and the circular section of the sphere in N.

Then, if the whole plane HFBC be conceived to revolve about the line HF as an axis, the square FG will describe a cylinder AG, and the quadrant IFG will describe a hemisphere EFG, and the triangle IFB will describe a cone IAB. Also, in the rotation, the three lines or parts кL, KN, Kм, as radii, will describe corresponding circular sections of those solids, name ly, KL a section of the cylinder, KN a section of the sphere, and Kм a section of the cone.

Now, FB being equal to F1 or IG, and KL parallel to FB, then by similar triangles IK is equal to KM (th. 82). And since, in the right-angled triangle IKN, IN2 is equal to IK2 + KN2 (th. 34); and because KL is equal to the radius IG or IN, and

KM

A

F

B

M

L

P

E

G

KM = IK, therefore I is equal to KM2 +KN2, or the square of the longest radius, of the said circular sections, is equal to the sum of the squares of the two others. And because circles are to each other as the squares of their diameters, or of their radii, therefore the circle described by KL is equal to both the circles described by Kм and KN; or the section of the cylinder, is equal to both the corresponding sections of the sphere and cone. And as this is always the case in every parallel position of KL, it follows, that the cylinder EB, which is composed of all the former sections, is equal to the hemisphere EG and cone IAB, which are composed of of all the latter sections.

D

H

But the cone IAB is a third part of the cylinder EB (cor. 2, th. 115); consequently the hemisphere EFG is equal to the remaining two-thirds; or the whole sphere EFGH equal to twothirds of the whole cylinder ABCD.

Q. E. D

Corol. 1. A cone, hemisphere, and cylinder of the same base and altitude, are to each other as the numbers 1, 2, 3.

Corol. 2. All spheres are to each other as the cubes of their diameters; all these being like parts of their circumscribing eylinders.

Corol. 3. From the foregoing demonstration it also appears, that the spherical zone or frustrum EGNP, is equal to the difference between the cylinder EGLO and the cone IMQ, all of the same common height IK. And that the spherical segment PFN, is equal to the difference between the cylinder ABLO and the conic frustrum AQMB, all of the same common altitude FK.

PROBLEMS.