drawn from A and E upon the planes BDC and FGH and let them be cut by planes parallel to BDC and FGH, and at equal altitudes QR and ST above those planes, and let the sections be the triangles KLM, NOP; KLM and NOP are equal to one another.

Because the plane ABD cuts the parallel planes BDC, KLM, the common sections BD and KM are parallel (14. 2. Sup.). For the same reason, DC and ML are parallel. Since therefore KM and ML are parallel to BD and DC, each to each, though not in the same plane with them, the angle KML is equal to the angle BDC (9. 2. Sup.). In like manner the other angles of these triangles are proved to be equal; therefore, the triangles are equiangular, and consequently similar; and the same is true of the triangles NOP, FGH.

:

Now, since the straight lines ARQ, AKB meet the parallel planes BDC and KML, they are cut by them proportionally (16. 2. Sup.), or QR: RA :: BK: KA; and AQ AR:: AB: AK (18. 5.), for the same reason, ES: ET: EF EN; therefore, AB : ÀK :: EF : EN, because AQ is equal to ES, and AR to ET. Again, because the triangles ABC, AKL are similar,

AB AK: BC: KL; and for the same reason

EF EN FG: NO; therefore,

BC: KL :: FG: NO. And, when four straight lines are proportionals, the similar figures described on them are also proportionals (22. 6.); therefore the triangle BCD is to the triangle KLM as the triangle FGH to the triangle NOP; but the triangles BDC, FGH are equal; therefore, the triangle KLM is also equal to the triangle NOP (1.5.). Therefore, &c. Q. E. D.

COR. 1. Because it has been shewn that the triangle KLM is similar to the base BCD; therefore, any section of a triangular pyramid parallel to the base, is a triangle similar to the base. And in the same manner it is shewn, that the sections parallel to the base of a polygonal pyramid are similar to the base.

COR. 2. Hence also, in polygonal pyramids of equal bases and alti

D d

tudes, the sections parallel to the bases, and at equal distances from them, are equal to one another.

PROP. XIII. THEOR.

A series of prisms of the same altitude may be circumscribed about any pyramid, such that the sum of the prisms shall exceed the pyramid by a solid less than any given solid.

H

N

W

V

M

Let ABCD be a pyramid and Z* a given solid; a series of prisms having all the same altitude, may be circumscribed about the pyramid ABCD, so that their sum shall exceed ABCD, by a solid less than Z. Let Z be equal to a prism standing on the same base with the pyramid, viz. the triangle BCD, and having for its altitude the perpendicular drawn from a certain point E in the line AC upon the plane BCD. It is evident, that CE multiplied by a certain numberm will be greater than AC; divide CA into as many equal parts as there are units in m, and let these be CF, FG, GH, HA, each of which will be less than CE. Through each of the points F, G, H let planes be made to pass parallel to the plane K BCD, making with the sides of the pyramid the sections FPQ, GRS, HTU, which will be all similar to one another, and to the base BCD (1. cor. 12. 3. Sup.). From the point B draw in the plane of the triangle ABC, the straight line BK parallel to CF meeting FP produced in K. In like manner, from D draw DL parallel to

B

G

m

E

包

S

then

CF, meeting FQ in L: Join KL, and it is plain, that the solid KBCDLF is a prism (def. 4. 3. Sup.). By the same construction, let the prisms PM, RO, TV be described. Also, let the straight line IP, which is in the plane of the triangle ABC be produced till it meet BC in h; and let the line MQ, be produced till it meet DC in g: Join bg; hC gQFP is a prism, and is equal to the prism PM (1. Cor. 8. 3. Sup.). In the same manner is described the prism MS equal to the prism RO, and the prism qU equal to the prism TV. The sum, therefore, of all the inscribed prisms hQ, mS, and qU is equal to the sum of the prisms PM, RO and TV, that is, to the sum of all the circumscribed prisms except the prism BL; wherefore, BL is the excess of the prisms circumscribed about the pyramid ABCD above the prisms inscribed with

The solid Z is not represented in the figure of this, or the following Proposition.

in it. But the prism BL is less than the prism which has the triangle BCD for its base, and for its altitude the perpendicular from E upon the plane BCD; and theprism which has BCD for its base, and the perpendicular from E for its altitude, is by hypothesis equal to the given solid Z; therefore, the excess of the circumscribed, above the inscribed prisms, is less than the given solid Z. But the excess of the circumscribed prisms above the inscribed is greater than their excess above the pyramid ABCD, because ABCD is greater than the sum of the inscribed prisms. Much more, therefore, is the excess of the circumscribed prisms above the pyramid, less than the solid Z. A series of prisms of the same altitude has therefore been circumscribed about the pyramid ABCD exceeding it by a solid less than the given solid Z. Q. E. D.

PROP. XIV. THEOR.

Pyramids 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. the perpendiculars drawn from the vertices A and E upon the planes, 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 solid 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 Z (13. 3. Sup.); let these be the prisms that have for their bases the triangles BCD, NQL, ORI, 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, UzX, VOY be made parallel to the base FGH. The section NQL is equal to the section WZT (12. 3. Sup.); as also ORI to XEU, and PSM to YoV; and therefore, also the prisms that stand upon the equal sections are equal (1. Cor. 8. 3. Sup.), that is, the prism which stands on the base BCD, and which is between the planes BCD and NQL is equal to the 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 prismsdescribed about the pyramid ABCD above the pyramid ABCD is less than Z (13. 3. Sup.); and therefore, the excess of the prisms described about the pyramid EFGH above the pyramid ABCD is also less than Z. But the excess of the pyramid EFGH above the pyramid ADCD is equal to Z, by hypothesis; therefore, the pyramid EFGH exceeds the pyramid ABCD, more than the prisms described about EFGH exceed the same pyramid ABCD. The pyramid EFGH is therefore greater than the sum of the prisms described about it, which is impossible. The pyramids ABCD, EFGH, therefore, are not unequal, that is, they are equal to one another. Therefore, pyramids, &c. Q. E. D.

PROP. XV. THEOR.

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

Let there be a prism of which the base is the triangle ABC, and let DEF be the triangle opposite 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 (34. 1.) to the triangle ABE therefore the pyramid of which the

base is the triangle ADE, and vertex the
point C, is equal (14. 3. Sup.) to the pyra-
mid, of which the base is the triangle ABE,
and vertex the point C. But the pyramid D
of which the base is the triangle ABE, and
vertex the point C, that is, the pyramid ABCE
is equal to the pyramid DEFC (14. 3. Sup.),
for they have equal bases, viz. the triangles
ABC, DEF, and the same altitude, viz. the
altitude of the prism ABCDEF. Therefore
the three pyramids ADEC, ABEC, DFEC
are equal to one another. But the pyra-
mids ADEC, ABEC, DFEC make up the
whole prism ABCDEF; therefore, the prism
ABCDEF is divided into three equal pyra-
mids. Wherefore, &c. Q. E. D.

A

C

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.

COR. 2. Pyramids of equal altitudes are to one another as their bases; because the prisms upon the same bases, and of the same altitude, are (1. Cor. 38. Sup.) to one another as their bases.

PROP. XVI. THEOR.

If 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 let AGHD be the rectangle (14. def. 3. Sup.) by the revolution of which the cylinder ABCD is described.

D

H

E

G

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 angles to the plane of the circle AEB. But EF is at right angles to the same plane; therefore, EF and GH are parallel A (6. 2. Sup.), and in the same plane. And since the plane through GH and EF cuts the parallel planes AEB, DFC, in the straight lines EG and FH, EG is parallel to FH (14. 2. Sup.). 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 AG. Therefore, when in the revolution of the rectangle AH, the straight line AG coincides with EG, the two rectangles AH and EH will coincide, and the straight line AD will coincide with the straight line EF. But AD is always in the superficies of the cylinder, for it describes that superficies; therefore, EF is also in the superficies of the cylinder. Therefore, &c. Q. E. D.