THEOREM CXIV.

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

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

Then, by the two preceding theorems,
DO2: DH2: LM: EF, and

AN2 AG2: IK : BC.

But since AN2, AG2, are equal to DO2, 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. È. 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.

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

Z 2

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 B, is the same solid as the pyramid whose base is BEF and vertex n, 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, BEF.

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 G (th. 41). Also, join CA, CB;

and

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

Then, because CG is perpendicular to the plane Adb, it is perpendicular both to GA and GD (def. 90). So that CGA, CGD are two right-angled triangles, having the perpendicular CG 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 G to the circumference of the section ADB, is equal to GA or GB; consequently that section is a circle.

Coral. 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.

For, let the plane AC be a section of the sphere and cylinder through the centre I. Join AI, BI. Also, let FIH be parallel to AD or BC, and EIG and KL parallel to AB or DC, the base of

E

M

the cylinder; the latter line KL meeting BI in M, 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 KL, KN, KM, as radii, will describe corresponding circular sections of those solids, namely, KL a section of the cylinder, KN a section of the sphere, and KM a section of the cone.

Now, FB being equal to FI 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 IK

KN2 (th. 34); and because KL is equal to the radius IG

or

or IN, and KмIK, therefore KL 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 be

Ε

D

MX

cause 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 EFG and cone IAB, which are composed of all the latter

sections.

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 two-thirds 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 cylinders.

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.

PROBLEM I.

To Bisect a Line AB; that is, to divide it into two Equal Parts.

FROM the two centres A and B, with any equal radii, describe arcs of circles, intersecting each other in c and D; and draw the line CD, which will bisect the given line AB in the point E.

For, draw the radii AC, BC, AD, BD. Then, because all these four radii are equal, and the side CD common, the two triangles

ACD, BCD, are mutually equilateral: consequently they are also mutually equiangular (th. 5), and have the angle AGE equal to the angle BCE.

Hence, the two triangles ACE, BCE, having the two sides AC, CE, equal to the two sides BC, CE, and their contained angles equal, are identical (th. 1), and therefore have the side AE equal to EB. Q. E. D.

PROBLEM II.

To Bisect an Angle BAC.

FROM the centre a, with any radius, describe an arc, cutting off the equal lines AD, AE; and from the two centres D, E, with the same radius, describe arcs intersecting in F; then draw AF, which will bisect the angle A as required.

For, join DF, EF. Then the two triangles ADF, AFF, having the two sides

AD, DF, equal to the two AE, EF (being equal radii), and the side AF common, they are mutually equilateral; consequently they are also mutually equiangular (th. 5), and have the angle BAF equal to the angle CAF.

Scholium. In the same manner is an arc of a circle bisected.

PROBLEM