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YRAMIDS that have equal bafes and altitudes
are equal to one another.

PR

Let ABCD, EFGH be two pyramids that have equal bafes BCD, FGH, and alfo equal altitudes, viz. the perpendiculars drawn from the vertices A and E upon the les BCD, FGH: The pyramid ABCD is equal to the pyramid EFGH.

If they are not equal, let the pyramid EFGH exceed the

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Book III.

pyramid ABCD by the folid Z. Then, a series of prisms of the fame altitude may be described about the pyramid ABCD that fhall exceed it, by a folid less than Za; let these be the prisms that have for their bases the triangles BCD, NQL, ORI, PSM. Divide EH into the fame number of equal parts into which AD is divided, viz. HT,TU, UV, VE, and through the points T, U and V, let the fections TZW, UZX, VOY be made parallel to the bafe FGH. The fection NQL is equal to the fection WZT b; as alfo ORI to XEU, and PSM to YOV; b 12.3.Sup. and therefore, also the prisms that stand upon the equal fections

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are

CI. cor. 38.
Sup.

Supplement are equal c, that is, the prifm which stands on the base BCD, and which is between the planes BCD and NQL is equal to the prifm which ftands on the bafe FGH, and which is between the planes FGH and WZT; and fo of the reft, because they have the fame altitude: wherefore, the fum of all the prifms defcribed about the pyramid ABCD is equal to the fum of all those described about the pyramid EFGH. But the excess of the prifms defcribed about the pyramid ABCD above

A

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the pyramid BCD is less than Z; and therefore, the excess
of the prifms defcribed about the pyramid EFGH above
the pyramid ABCD is also less than Z. But the excess of
the pyramid EFGH above the pyramid ABCD is equal to
Z, by hypothefis; therefore, the pyramid EFGH exceeds
the pyramid ABCD, more than the prifms defcribed about
EFGH exceed the fame pyramid ABCD. The pyramid
EFGH is therefore greater than the fum of the prisms de-
fcribed about it, which is impoffible. The pyramids ABCD,
EFGH, therefore, are not unequal, that is, they are equal to
one another. Therefore, pyramids, &c. Q. E. D.

PROP

Book III.

E

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VERY prism having a triangular base may be divided into three pyramids that have triangular bafes, and that are equal to one another.

Let there be a prifm of which the bafe is the triangle ABC, and let DEF be the triangle oppofite to the base: The prism ABCDEF may be divided into three equal pyramids having triangular bafes.

D

F

E

b 14.3.Sup.

Join AE, EC, CD; and because ABED is a parallelogram, of which AE is the diameter, the triangle ADE is equal a to the triangle ABE: therefore the pyramid of which a. 34. I. the base is the triangle ADE, and vertex the point C, is equal b to the pyramid, of which the bafe is the triangle ABE, and vertex. the point C. But the pyramid of which the base is the triangle ABE, and vertex the point C, that is, the pyramid ABCE is equal to the pyramid DEFCf, for they have equal bases, viz. the triangles ABC, DFE, and the fame altitude, viz. the altitude of the prifm ABCDEF. Therefore, the three pyramids ADEC, ABEC, DFEC are equal to one another. But the pyramids ADEC, ABEC, DFEC make up the whole prism ABCDEF; therefore, the

A

B

prifm ABCDEF is divided into three equal pyramids. Wherefore, &c. Q. E. D.

COR. 1. From this it is manifeft, that every pyramid is the third part of a prifm which has the fame bafe, and the fame altitude with it; for if the bafe of the prifm be any other figure than a triangle, it may be divided into prisms having triangular bafes.

COR. 2. Pyramids of equal altitudes are to one another as their bafes; because the prisms upon the fame bases, and of the fame altitude, are to one another as their bases.

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PROP,

Supplement

PROP. XVI.

THEOR.

IF

F from any point in the circumference of the base of a cylinder, a ftraight line be drawn perpendicular to the plane of the bafe, it will be wholly in the cylindric fuperficies.

Let ABCD be a cylinder, of which the bafe is the circle AEB, DFC the circle oppofite to the bafe, and GH the axis; from E, any point in the circumference AEB, let EF be drawn perpendicular to the plane of the circle AEB; the straight line EF is in the fuperficies of the cylinder.

Let F be the point in which EF meets the plane DFC oppofite to the bafe; join EG and FH; and a 14. def. 3. let AGHD be the rectangle by the revolution of which the cylinder ABCD is described.

Sup.

Now, because GH is at right angles to GA, the ftraight line. which by its revolution describes the circle AEB, it is at right angles to all the ftraight lines in the plane of that circle which meet it in G, and it is therefore at right angles to the plane of the circle AEB. But EF is at right angles to the fame plane; therefore, EF

D

A

E

H

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b 6. 2. Sup. and GH are parallel b, and in the fame plane. And fince the plane through GH and EF cuts the parallel planes AEB, 14. 2. Sup. DFC in the ftraight lines EG and FH, EG is parallel to FH. The figure EGHF is therefore a parallelogram, and it has the angle EGH a right angle, therefore it is a rectangle, and is equal to the rectangle AH, becaufe EG is equal to AG. Therefore, when in the revolution of the rectangle AH, the ftraight line AG coincides with EG, the two rectangles AH and EH will coincide, and the ftraight line AD will coin

cide

cide with the ftraight line EF. But AD is always in the Book III. fuperficies of the cylinder, for it defcribes that fuperficies; therefore, EF is alfo in the fuperficies of the cylinder. Therefore, &c. Q. E. D.

A

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CYLINDER and a parallelepiped having equal
bafes and altitudes, are equal to one another.

Let ABCD be a cylinder, and EF a parallelepiped having equal bases, viz. the circle AGB and the parallelogram EH, and having also equal altitudes; the cylinder ABCD is equal to the parallelepiped EF.

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If not, let them be unequal; and firft, let the cylinder be lefs than the parallelepiped EF; and from the parallelepiped EF let there be cut off a part EQ by a plane PQ parallel to NF, equal to the cylinder ABCD. In the circle AGB infcribe the polygon AGKBLM that shall differ from the circle by a space lefs than the parallelogram PH a, and cut off from a cor. I. 4. 1, Sup. the parallelogram EH, a part OR equal to the polygon AGKBLM. The point R will fall between P and N. On the polygon AGKBLM let an upright prifm AGBCD be conftituted of the fame altitude with the cylinder, which will

therefore

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