PROPOSITION III.

THEOREM. In every prism, the sections formed by parallel planes are equal polygons.

In the prism ABCI, let the sections NOPQR, STVXY be formed by parallel planes; then will these sections be equal polygons.

Y

G

V

R

B

H

For the sides ST, NO are parallel, being the intersections of two parallel planes with a third plane ABGF; moreover the sides ST, NO are included between the parallels NS, OT, which are sides of the prism hence NO is equal to ST. For like reasons, the sides OP, PQ, QR, &c. of the section NOPQR, are respectively equal to the sides TV, VX, XY, &c. of the section STVXY; and since the equal sides are at the same time parallel, it follows that the angles NOP, OPQ, &c. of the first section are respectively equal to the angles STV, TVX, &c. of the second hence the two sections NOPQR, STVXY are equal polygons.

Cor. Every section in a prism, if made parallel to the base, is also equal to that base.

PROPOSITION IV.

THEOREM. If a plane be made to pass through the diagonal and opposite edges of a parallelopipedon, so as to divide it into two triangular prisms, those prisms are equal.

Let the parallelopipedon ABCG

be divided by the plane BDHF into the two triangular prisms ABDHEF, BCDFGH; then will those prisms be equal.

E

H

D

g

then the sec

Through the vertices B and F, draw the planes Badc, Fehg, at right angles to the side BF, and meeting AE, DH, CG, the three other sides of the parallelopipedon, in the points a, d, c towards one direction, and in e, h, g towards the other tions Badc, Fehg will be equal parallelograms; being equal, because they are formed by planes perpendicular to the same straight line, and consequently parallel; and being parallelograms, because aB, dc, two opposite sides. of the same section, are formed by the meeting of one plane with two parallel planes ABFE, DCGH.

For a like reason, the figure BaeF is a parallelogram ; so also are BF gc, cdhg and adhe, the other lateral faces of the solid BadcFehg: hence that solid is a prism (Def. 4); and that prism is a right one, because the side BF is perpendicular to its base.

This being proved, if the right prism Bh be divided by the plane BFHD into two right triangular prisms aBdeFh,

BdcFhg, it will remain to be shown that the oblique triangular prism ABDEFH will be equal to the right triangular prism aBdeFh; and since those two prisms have a part ABDheF in common, it will only be requisite to prove that the remaining parts, namely, the solids BaADd, FeEHh, are equal.

Now, by reason of the parallelograms ABFE, aBFe, the sides AE, ae, being equal to their parallel BF, are equal to each other; and taking away the common part Ae, there remains Aa=Ee. In the same manner we could prove Dd=Hh.

Let us now place the base Feh on its equal Bad; the point e coinciding with a, and the point h with d, the sides eE, hH will coincide with their equals aA, dD, because they are perpendicular to the same plane Bad. Hence the two solids in question will coincide exactly with each other, and the oblique prism BADFEH is therefore equal to the right one BadFeh.

In the same manner might the oblique prism BDCFHG be proved equal to the right prism BdcFhg. But (B. VII, Prop. 1) the two right prisms BadFeh, BdcFhg are equal, since they have the same altitude BF, and since their bases Bad, Bdc are halves of the same parallelogram. Hence the two triangular prisms BADFEH, BDCFHG, being equal to the equal oblique prisms, are equal to each other.

Cor. Every triangular prism ABDHEF is half of the parallelopipedon AG described on the same solid angle A, with the same edges AB, AD, AE.

PROPOSITION V.

L

M G

H

IF

THEOREM. Two parallelopipedons having a common base, and their upper bases in the same plane and between the same parallels, are equal to each other. Let the two parallelopipedons AG, AL have the common base ABCD; and let their upper bases EFGH, IKLM be in the same plane, and between the same parallels EK, HL then will these parallelopipedons be equal.

K

B

There may be three cases to this proposition, according as EI is greater, less than, or equal to EF; but the demonstration is the same for all. In the first place, then, we shall show that the triangular prism AEIDHM is equal to the triangular prism BFKCGL.

Since AE is parallel to BF, and HE to GF, the angle AEI=BFK, HEI=GFK, and HEA=GFB. Of these six angles, the first three form the solid angle E, and the last three the solid angle F; therefore, the plane angles being respectively equal and similarly arranged, the solid angles F and E must be equal. Now, if the prism AEM be laid on the prism BFL, the base AEI, being placed on the base BFK, will coincide with it, because they are equal; and since the solid angle E is equal to the solid angle F, the side EH will coincide with its equal FG; and nothing more is required to prove the coincidence of the two prisms throughout their whole extent, for (B. VII, Prop. 1) the base AEI and the edge EH determine the prism AEM, as the base BFK and the edge FG determine the prism BFL: hence these prisms are equal.

But if the prism AEM is taken away from the solid AL, there will remain the parallelopipedon AIL; and if the prism BFL is taken away from the same solid, there will remain the parallelopipedon AEG hence these two parallelopipedons AIL, AEG are equal.

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PROPOSITION VI.

THEOREM. Two parallelopipedons having the same base and same altitude, are equal to each other.

Let ABCD be the common base of the two parallelopipedons AG, AL : since they have the same altitude, their upper bases EFGH, IKLM will be in the same plane. Also the sides EF and AB will be equal and parallel, as well as IK and AB; hence EF is equal and parallel to IK for a like reason, GF is equal and parallel to LK. Let the sides EF, HG be produced, and likewise LK, IM, till, by their intersections, they form the parallelogram NOPQ this parallelogram will evidently be equal to

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