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YRAMIDS that have equal bases and altitudes
are equal to one another.
Let ABCD, EFGH be two pyramids that have equal bases BCD, FGH, and also equal altitudes, viz. tlie perpendiculars drawn from the vertices A and E upon th: il es BCD, FGH: The pyramid ABCD is equal to the pyramid EFGH.
If they are not equal, let the pyramid EFGH exceed the
pyramid ABCD by the folid Z. Then, a series of prisms of the same altitude may be described about the pyramid ABCD that shall exceed it, by a solid less than Za; let these be the prisms that have for their bases the triangles BCD, NQL, ORĪ, PSM. Divide EH into the same number of equal parts into which AD is divided, viz. HT, TU,UV,VE, and through the points T, U and V, let the sections TZW,UEX, VOY be made parallel to the base FGH. The section NQL is equal to the section WZTb; as also ORI to X.EU, and PSM to YOV; b 12.3.Sup. and therefore, also the prisms that stand upon the equal sections
Supplement are equal c, that is, the prism which stands on the base BCD,
and which is between the planes BCD and MQL is equal to the C1. cor. 38. Sup. prism which stands on the base FGH, and which is between the
planes FGH and WZT, and so of the rest, because they have the same altitude : wherefore, the sum of all the prisms described about the pyramid ABCD is equal to the sum of all those described about the pyramid EFGH. But the excess of the 'prisins described about the pyramid ABCD above
G the pyramid BCD is less than Z; and therefore, the excess of the prisms described about the pyramid EFGH above the pyramid ABCD is also less than 2. But the excess of the pyramid EFGH above the pyramid ABCD is equal to 2, by hypothesis ; therefore, the pyramid EFGH exceeds the pyramid ABCD, more than the prisms described about EFGH exceed the lame pyramid ABCD. The pyramid EFGH is therefore greater than the sum of the prisms described about it, which is impoflible. The pyramids ABCD, EFGH, therefore, are not unequal, that is, they are equal to one another. Therefore, pyramids, &c. Q. E. D.
VERY prism having a triangular base may be
divided into three pyramids that have triangular bases, and that are equal to one another.
Let there be a prism of which the base is the triangle ABC, and let DEF be the triangle opposite to the base : The prism ABCDEF may be divided into three equal pyramids having triangular bases.
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 2. 34. 8. the base is the triangle ADE, and vertex the point C, is equal b to the pyramid, of which the
b 14.3. Sup. base is the triangle ABE, and vertex the point C. But the pyramid of which the base is the triangle ABE,
E 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 same altitude, viz. the alti. tude of the prism ABCDEF. Therefore, the three pyramids ADEC, ABEC, DFEC are equal to one a
С nother. But the pyramids ADEC, ABEC, DFEC make up the whole
*B prism ABCDEF; therefore, the prism ABCDEF is divided into three equal pyramids. Wherefore, &c. Q. E. D.
Cor. 1. From this it is manifest, that every pyramid is the third part of a prism which has the same base, and the same altitude with it; for if the base of the prism be any other figure than a triangle, it may be divided into prisms having triangular bases.
Pyramids of equal altitudes are to one another as their bases; because the prisms upon the same bases, and of the same altitude, are c to one another as their bases. CI. cor. 38
F from any point in the circumference of the base
of a cylinder, a straight line be drawn perpendicular to the plane of the base, it will be wholly in the cylindric superficies.
Let ABCD be a cylinder, of which the base is the circle AEB, DFC the circle opposite to the base, 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 superficies of the cylinder.
Let F be the point in which EF meets the plane DFC opposite to the base ; join EG and FH; and
ABCD is described.
Now, because GH is at right angles to GA, the straight line which by its revolution describes the circle AEB, it is at right angles to all the straight lines in the plane of that circle which meet it in G, and it is therefore at right
to the same plane ; therefore, EF 8 6. 2. Sup. and GH are parallel b, and in the same plane. And since the
plane through GH and EF cuts the parallel planes AEB, ¢14. 2. Sup. DFC in the straight lines EG and FH, EG is parallel to FHS.
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, because EG is equal to AĢ. Therefore, when in the revolution of the rectangle AH, the straight line AG coincides with EG, the two rečtangles AH and EH will coincide, and the straight line AD will coin.
cide with the straight line EF. But AD is always in the Book III. superficies of the cylinder, for it describes that fuperficies; therefore, EF is also in the superficies of the cylinder. Therefore, &c. Q. E. D.
CYLINDER and a parallelepiped having equal
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.
If not, let them be unequal ; and first, let the cylinder be less 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 less than the parallelogram PH a, and cut off from a cor. 1. 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 prism AGBCD be copstituted of the fame altitude with the cylinder, which will