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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 equivalent to the equal oblique prisms, are equivalent to each other.

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

PROPOSITION IX. THEOREM.

If two parallelepipedons AG, AL have a common base ABCD, and if their upper bases EFGH, IKLM lie in the same plane and between the same parallels EK, HL, those two parallelepipedons will be equivalent.

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There be three cases, may 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 shew that the triangular prism D 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, 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 is 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 fall on its equal FG: and nothing more is required to prove the coincidence of the two prisms throughout their whole extent, for (3. VI.) 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 parallelepipedon AIL; and if the prism BFL is taken away from the same solid, there will remain the parallelepipedon AEG; hence those two parallelepipedons AIL, AEG are equivalent.

PROPOSITION X. THEOREM.

Two parallelepipedons having the same base and the same altitude are equivalent.

Let ABCD (see the next figure) be the common base of the two parallelepipedons 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 either of the bases EFGH, ÏKLM. Now if a third parallelepipedon be conceived, having for its lower base the same ABCD, and NOPQ for its upper, this third parallelepipedon will (9. VI.) be equal to the parallelepipedon AG, since with the same lower base, their upper bases lie in the same plane and between the same parallels GQ, FN. For the same reason, this third parallelepipedon will also be equivalent to the parallelepipedon AL; hence the two parallelepipedons AG, AL, which have the same base and the same altitude, are equivalent.

PROPOSITION XI. THEOREM.

Any parallelepipedon may be changed into an equivalent rectan gular parallelepipedon having the same altitude and an equivalent base.

Let AG be the parallele- q pipedon proposed. From the points A, B, C, D, N draw AI, BK, CL, DM, perpendicular to the plane of the base; you will thus form the parallelepipedon AL equivalent to AG, and having its lateral faces AK, BL, &c. rectangular. Hence if the base ABCD is a rectangle, AL will be the rect

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equivalent to AG, the parallelepipedon proposed. But if ABCD

is not a rectangle, draw AO and BN perpendicular to CD, and OQ and NP perpendicu- MQ lar to the base; you will then have the solid ABNOIPQ, which will be a rectangular parallelepipedon: for by construction, the base ABNO and its opposite IKPQ are rectangles; so also are the lateral faces, the edges AI, OQ, &c. being perpendicular to the plane of the base; hence the solid AP is a rectangular parallelepipedon. But the two parallelepipedons AP, AL may be conceived as having the same base ABKI and the same altitude AO: hence the parallelepipedon AG, which was at first changed into an equivalent parallelepipedon AL, is again changed into an equivalent rectangular parallelepipedon AP, having the same altitude AI, and a base ABNO equivalent to the base ABCD.

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PROPOSITION XII. THEOREM.

Two rectangular parallelepipedons AG, AL, which have the same base ABCD, are to each other as their altitudes AE, AI.

First, suppose the altitudes AE, AI, to be E to each other as two whole numbers, as 15 is to 8, for example. Divide AE into 15 equal parts; whereof AI will contain 8; and through x, y, z, &c. the points of division, draw planes parallel to the base. These planes will cut the solid AG into 15 partial parallelepipedons, all equal to each other, because having equal bases and equal altitudes,-equal bases, since (7. VI.) every section MIKL, made parallel to the base ABCD of a prism, is equal to that base,-equal altitudes because these altitudes are the same divisions Ax, xy, xy, &c. But of those 15 equal parallelepipedons, & are contained in AL; hence the solid AG is to the solid AL as 15 is to 8, or generally, as the altitude AE is to the altitude AI..

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Again, if the ratio of AE to AI cannot be expressed in numbers, it is to be shewn, that notwithstanding, we shall have solid. AG: solid. AL:: AE: AI. For, if this proportion is not correct, suppose we have sol. AG: sol. AL:: AE: AO. Divide AE into equal parts, such that each shall be less than OI; there will be at least one point of division m between O and I. Let P be the parallelepipedon, whose base is ABCD, and alti

tude Am; since the altitudes AE, Am are to each other as two whole numbers, we shall have sol. AG: P:: AE: Am. But by hypothesis, we have sol. AG: sol. AL:: AE: AO; therefore sol. AL: P:: AO: Am. But AO is greater than Am; hence if the proportion is correct, the solid AL must be greater than P. On the contrary, however, it is less hence the fourth term of this proportion sol. AG: sol. AL:: AE: x, cannot possibly be a line greater than AI. By the same mode of reasoning, it might be shewn that the fourth term cannot be less than AI; therefore it is equal to AI: hence rectangular parallelepipedons having the same base are to each other as their altitudes.

PROPOSITION XIII. THEOREM.

Two rectangular parallelepipedons AG, AK, having the same altitude AE, are to each other as their bases BBCD, AMNO.

Having placed the two I solids by the side of each other, as the figure represents, produce the plane ONKL till it meets the plane DCGH in PQ; you will thus have a third parallelepipedon AQ, which may be compared with each of the parallelepipedons AG, AK. The two solids AG, AQ, having the same base AEHD, are to each other as their altitudes AB, AO; in like manner, the two solids AQ, AK, having the same base AOLE, are to each other as their al

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titudes AD, AM. Hence we have the two proportions, sol. AG: sol. AQ:: AB: AO,

sol. AQ sol. AK:: AD: AM.

Multiply together the corresponding terms of those proportions, omitting in the result the common multiplier sol. AQ; we shall have

sol. AG sol. AK:: ABXAD: AOXAM.

But ABX AD represents the base ABCD; and AOX AM represents the base AMNO: hence two rectangular parallelepipedons of the same altitude are to each other as their bases.

PROPOSITION XIV. THEOREM.

Any two rectangular parallelepipedons are to each other as the products of their bases by their altitudes, that is to say, as the products of their three dimensions.

For, having placed the two solids AG, AZ, so that their surfaces have the common angle BAE, produce the interior planes necessary for completing the third parallelepipedon AK, hav

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ing the same altitude with the parallelepipedon AG. By the last Proposition, we shall have

sol. AG: sol. AK :: ABCD: AMNO.

But the two parallelepipedons AK, AZ having the same base AMNO, are to each other as their altitudes AE, AX; hence we have

sol. AK: sol. AZ:: AE: AX.

Multiply together the corresponding terms of those proportions, omitting in the result the common multiplier sol. AK; we shall have

sol. AG: sol. AZ:: ABCDXAE: AMNO XAX. Instead of the bases ABCD and AMNO, put ABXAD and AOXAM; it will give

sol. AG: sol. AŽ :: AB× ADX AE: AO×AM XAX. Hence any two rectangular parallelepipedons are to each other, &c.

Scholium. We are consequently authorised to assume, as the measure of a rectangular parallelepipedon, the product of

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