solidity of the latter is equal to its base multiplied by its height: hence the solidity of the former is, in like manner, equal to the product of its base by its altitude.

In the second place, and for a like reason, any rectangular prism is half of the parallelopipedon so constructed as to have the same altitude and a double base; but the solidity of the latter is equal to its base multiplied by its altitude: hence that of a triangular prism is also equal to the product of its base multiplied into its altitude.

In the third place, any prism may be divided into as many triangular prisms of the same altitude, as there are triangles capable of being formed in the polygon which constitutes its base: but the solidity of each triangular prism is equal to its base multiplied by its altitude; and since the altitude is the same for all, it follows that the sum of all the partial prisms must be equal to the sum of all the partial triangles which constitute their bases, multiplied by the common altitude.

Hence the solidity of any polygonal prism is equal to the product of its base by its altitude.

Cor. Comparing two prisms which have the same altitude, the products of their bases by their altitudes will be as the bases simply: hence two prisms of the same altitude are to each other as their bases. For a like reason, two prisms of the same base are to each other as their altitudes.

PROPOSITION XII.

THEOREM. Similar prisms are to one another as the cube of their homologous sides.

From A and a, homologous angles of the two prisms, draw AH, ah perpendicular to the bases BCD, bcd: join BH; take Ba=ba, and in the plane BHA draw ah perpendicular to BH: then ah shall be perpendicular to the plane CBD, and equal to ah the altitude of the other prism; for if the solid angles B and b were applied the one to the other, the planes which contain them, and consequently the perpendiculars ah, ah would coincide. Now because of the similar triangles ABH, abh, and the similar figures AC, ac, we have

AH ah AB ab: BC: bc;

and because of these similar bases, the

base BCD base bcd: BC: bc'. [B. IV, Prop. xxi.] Taking the product of the corresponding terms of these proportions, we have

AH x base BCD: ahx base bcd: BC3: bc3.

But AH base BCD expresses the solidity of the prism

P, and ah × base bcd expresses the solidity of the other prism p; therefore

prism P prism p: BC: bc3.

PROPOSITION XIII.

THEOREM. The section formed by cutting a triangular pyramid by a plane parallel to the base, is similar to the base.

Let the triangular pyramid ABCD be cut by a plane parallel to the base, forming the section FGH; then will this section be similar to the base BCD.

For because the parallel planes BCD, FGH are cut by a third plane

D

K

E

B

ABC, the sections FG, BC are parallel. (B. VI, Prop. x.) In like manner, it appears that FH is parallel to BD; therefore the angle HFG is equal to the angle DBC, (B. VI, Prop. XIII ;) and because the triangle ABC is similar to the triangle AFG, and the triangle ABD is similar to the triangle AFH, we have

BC: BA :: FG : FA, and

BA: BD :: FA: FH; therefore

BC: BD :: FG: FH.

Now the angle DBC has been shown to be equal to the angle HFG; therefore the triangles DBC, HFG are equiangular, (B. IV, Prop. xiv.)

19*

PROPOSITION XIV.

THEOREM. If two triangular pyramids, which have equal bases and equal altitudes, be cut by planes which are parallel to the bases, and at equal distances from them; those sections will be equal.

equal distances, forming the sections FGH, fgh; then will these sections be equal.

Draw AKE, ake perpendicular to the bases BCD, bcd, meeting the cutting planes in K and k; then, because of the parallel planes, we have, (B. VI, Prop. xix,)

AE : AK :: AB : AF, and ae: ak :: ab: af; but, by hypothesis, AE=ae, and AK=ak; therefore AB AF: ab: af.

:

Again, because of similar triangles,

AB : AF :: BC : FG, and ab : af :: bc : fg ; and, hence, BC: FG :: bc: fg";

but, because of the similar triangles BDC, FGH,

:

BC FG triangle BDC: triangle FHG. And, in like manner,

bcfg trian. BCD

triangle bdc

triangle fgh; therefore

trian. FGH

trian. bcd trian. fgh

Now, trian. BCD=trian. bcd, (by hypothesis ;) therefore the triangle FHG is equal to the triangle fhg.

Scholium. It is easy to see, that what is proved in this and the preceding proposition, is also true of polygonal pyramids.

PROPOSITION XV.

THEOREM. Two triangular pyramids having equivalent bases and equal altitudes, are equivalent, or equal in solidity.

Let SABC, Sabc be those two pyramids; let their equivalent bases ABC, abc be situated in the same plane, and let AT be their common altitude. If they are not equivalent, let Sabc be the smaller; and suppose Aa to