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gram KL is similar and equal to the parallelogram CN. For the saine reason the parallelogram MK is similar and equal to the parallelogram CR; and also OE to FD. Therefore three parallelograms of the solid KO are equal and similar to three parallelograms of the solid CD. But the three opposite ones in each solid are equal (XI. 24) and similar to these. Therefore the solid K'O is equal (XI. C) and similar to the solid CD. Complete the parallelogram GK; and upon the bases GK and KL complete the solids EX and

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LP, so that EH be an insisting straight line in each of them, and that they are of the same altitude with the solid A B.

Because the solids A B and C D are similar, AE is to E G as CF is to F N. Therefore, by permutation, A E is to CF, as E G is to FN, and as EH is to FR. But FC is equal to EK, FN to EL, and FR to EM. Therefore A E is to EK, as E G is to E L, and as HE is to EM. But A E is to EK (VI. 1), as the parallelogram AG is to the paralleloGK; gram and GE is to EL, as (VI. 1) GK is to KL; and HE is to EM (VI. 1), as PE is to K M. Therefore the parallelogram A G is to the parallelogram G K, as GK is to KL, and as PE is to KM. But AG is to G K, as (XI. 25) the solid A B is to the solid EX: and GK is to KL (XI. 25), as the solid EX is to the solid PL; and PE is to K M, as (XI. 25) the solid PL is to the solid KO. Therefore the solid A B is to the solid EX, as EX is to P L, and as PL is to KO. But if four magnitudes be continual proportionals, the first is said to have to the fourth the triplicate (V. Def. 11) ratio of that which it has to the second. Therefore the solid A B has to the solid K O, the triplicate ratio of that which AB has to EX. But A B is to EX, as the parallelogram A G is to the parallelogram GK, and as the straight line AE is to the straight line E K. Therefore the solid AB has to the solid KO the triplicate ratio of that which AE has to EK. But the solid KO is equal to the solid CD, and the straight line EK is equal to the straight line CF. Therefore the solid AB has to the solid CD the triplicate ratio of that which the side AE has to the homologous side CF. Therefore, similar parallelopipeds, &c. Q. E. D.

COROLLARY.-From this it is manifest that, if four straight lines be continual proportionals, as the first is to the fourth, so is the parallelopiped described from the first as a side or edge to the similar solid similarly described from the second as a side or edge; because the first straight line has to the fourth the triplicate ratio of that which it has to the second.

This proposition is one of the most important in this Book. It is the foundation of the rule for determining the solid content of all similar solids bounded by planes.

PROP. D. THEOREM.

Parallelopipeds which are contained by parallelograms equiangular to one another, each to each, that is, of which the solid angles are equal, each to each, have to one another the ratio which is the same with the ratio compounded of the ratios of their sides.

Let A B and CD be parallelopipeds, of which A B is contained by the parallelograms AE, AF, and AG, equiangular, each to each, to the parallelograms CH, CK, and CL, which contain the solid CD. The ratio which the solid A B has to the solid CD, is the same with that which is compounded of the ratios of the sides AM to DL, AN to DK, and AO to DH.

Produce MA, NA, and OA to P, Q, and R, making AP equal to DL, AQ to DK, and AR to DH. Complete the parallelopiped AX contained by the parallelograms AS, AT, and AV, similar and equal to CH, CK, and CL, each to each. The solid AX is equal (XI. C) to the solid CD. Complete likewise the solid A Y, of which the base is AS, and AO one of its insisting straight lines. Take any straight line a, and as M A is to AP, so make (VI. 12) a to b; and as NA is to AQ, so make b to c; and as AO is to AR, so make c to d.

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Because the parallelogram AE is equiangular to A S, A E is to AS as a is to c (VI. 23). But the solids A B and AY, between the parallel planes BOY and E A S, are of the same altitude. Therefore the solid A B is to the solid A Y as (XI. 32) the base AE is to the base AS; that is, as a is to c. But the solid A Y is to the solid A X as (XI. 25) the base OQ is to the base Q R; that is, as O A is to A R; that is, as c is to d. Because the solid A B is to the solid A Y as a is to c, and the solid A Y is to the solid A X D as c is to d; ex æquali the solid AB is to the solid A X, or CD, H which is equal to it, as a is to d. But the ratio of a to d is said to be compounded (V. Def. A) ofa the ratios of a to b, b to c, and c to d, which are the same with the ratios of the sides M A to A P, NA to A Q, and O A to A R, each to each and the sides, AP, AQ and AR are equal to the sides DL, DK, and D H, each to each. Therefore the solid A В has to the solid CD the ratio which is the same with that which is compounded of the ratios of the sides AM to DL, AN to DK, and AO to DH. Q. E. D. DE This proposition was introduced into this Book by Dr. Simson.

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C d

PROP. XXXIV. THEOREM.

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The bases and altitudes of two equal parallelopipeds, are reciprocally proportional: and conversely, if the bases and altitudes of two parallelopipeds be reciprocally proportional, they are equal.

Let A B and CD be two equal parallelopipeds. altitudes are reciprocally proportional.

Their bases and

First, let their insisting straight lines AG, EF, LB, and HK; CM, NX, OD, and PR be at right angles to their bases. The base EH is to the base NP as CM is to AG.

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If the base EH be equal to the base NP; because the solid A B is (Hyp.) equal to the solid CD, CM is equal to AG. For, if the bases EH and NP be equal, but the altitudes AG and CM not equal, neither is the solid AB equal to the solid CD. But the solids (Hyp.) are equal. Therefore the altitude CM is not unequal to the altitude A G; that is, CM is equal to A G. Wherefore the base EH is to the base NP, as CM is to A G. If the bases EH and NP be not equal, let E H be greater than the other. Because the solid AB is equal to the solid ČD, CM is greater than A G. For, if not, the solids AB and CD are unequal. But (Hyp.) they are equal. Make C'Î equal to CT A G, and complete the parallelopiped CV, of which the base is N P, and altitude CT. Because the solid AB is equal to the solid CD, the solid A B is to the solid C V, as (V. 7) the H solid CD is to the solid CV. But the solid AB is to the solid CV, as (XI. 32) the base EH is to the base NP; for the solids A B and CV are of the same altitude. Because the solid CD is to the solid C V, as (XI. 25) the base MP is to the base P T, and as (VI. 1) MC is to CT; and CT is equal to A G. Therefore the base EH is to the base NP, as MC is to A G. Wherefore the bases and altitudes of the parallelopipeds A B and CD are reciprocally proportional.

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IV

Conversely, let the bases and altitudes of the parallelopipeds A B and CD be reciprocally proportional. They are equal.

First, let their insisting straight lines be at right angles to their bases; and let the base EH be to the base NP, as CM is to AG. Let the base EH be equal to the base NP. Because EH is to NP, as the altitude of the solid CD is to the altitude of the solid A B. Therefore the altitude of CD is equal (V. A) to the altitude of A B. But parallelopipeds upon equal bases, and of the same altitude, are equal (XI. 31) to one another. Therefore the solid A B is equal to the solid CD. But let the bases E H and N P be unequal, and let E H be the greater. Because the base EH is to the base NP, as CM, the altitude of the solid CD, is to AG the altitude of AB, CM is greater (V. A) than AG. Make CT equal to AG, and complete the solid CV. Because the base EH is to the base NP, as CM is to A G, and AG is equal to CT. Therefore the base EH is to the base NP, as MC to CT. But the base E H is to the base NP, (XI. 32) as the solid A B is to the solid CV; for the solids A B and C V are of the same altitude: and M C is to CT, as (VI. 1) the base M P is to the base P T, and as the solid CD is to the solid (XI. 25) CV. Therefore the solid A B is to the solid CV, as the solid CD is to the solid CV; that is, each of the solids A B and CD) has the same ratio to the solid CV. Therefore the solid A B is equal (V. 9) to the solid CD.

Secondly, let their insisting straight lines FE, BL, GA, and K H

XN, DO, MC, and RP be not at right angles to their bases. The base EH is to the base N P, as the altitude of CD is to the altitude of A B. From the points F, B, K, and G; X. D, R, and M, draw perpendiculars to the planes in which are the bases EH and N P, meeting those planes in the points S, Y, V, and T; Q, I, U, and Z; and complete the solids FV and X U, which are parallelopipeds (XI. 31).

K B

Because the solid AB is equal to the solid CD, and the solid A B is equal (XI. 29 or 30) to the solid BT, for they are upon the same base FK, and of the same altitude; and the solid CD is equal (XI. 29 or 30) to the solid DZ, being upon the same base XR, and of the same altitude. Therefore the solid BT is equal to the solid DZ. But the bases and altitudes of equal parallelopipeds of which the insisting straight lines are at right angles H to their bases, are reciprocally proportional. Therefore the base FK is to the base XR, as the altitude

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of the solid DZ is to the altitude of the solid BT: and the base FK is equal to the base EH, and the base X R to the base N P. Therefore, the base EH is to the base NP, as the altitude of the solid D Z is to the altitude of the solid BT: but the altitudes of the solids DZ and DC, as also of the solids B T and B A, are the same. Therefore the base EH is to the base N P, as the altitude of the solid CD is to the altitude of the solid A B; that is, the bases and altitudes of the parallelopipeds A B and CD are reciprocally proportional.

Conversely, let the bases of the solids A B and CD be reciprocally proportional to their altitudes, viz., the base EH to the base NP, as the altitude of the solid CD to the altitude of the solid A B. The solid A B shall be equal to the solid CD.

The same construction being made; because the base EH is to the base NP, as the altitude of the solid CD is to the altitude of the solid AB; and the base EH is equal to the base FK, and NP to XR. Therefore the base FK is to the base X R, as the altitude of the solid CD to the altitude of A B. But the altitudes of the solids A B and B T are the same, as also of CD and D Z. Therefore the base F K is to the base X R, as the altitude of the solid DZ is to the altitude of the solid BT. Wherefore the bases of the solids BT and DZ are reciprocally proportional to the altitudes: and their insisting straight lines are at right angles to their bases. Therefore, as was before proved, the solid BT is equal to the solid D Z. But BT is equal (XI. 29 or 30) to B A, and D Z to DC, because they are upon the same bases, and of the same altitude. Therefore the solid A B is equal to the solid CD. Therefore the bases, &c. Q. E. D.

PROP. XXXV. THEOREM.

If, from the vertices of two equal plane angles, there be drawn two straight lines elevated above the planes in which the angles are, and containing equal angles with the sides of those angles, each to each; and if in the lines above the planes there be taken any points, and from them perpendiculars be drawn to those planes; and if from the points in which they meet the planes, straight lines be drawn to the vertices of the two equal plane angles; these straight lines make equal angles with the straight lines above the planes.

Let B A C and EDF be two equal plane angles; and from the points A and D let the straight lines AG and DM be drawn above the planes of the angles, making equal angles with their sides, viz., GAB equal to MDE, and G A C to MDF. From G and M, any points in the straight lines AG and DM, let perpendiculars GL and MN be drawn (XI. 11) to the planes BAC and EDF meeting them in the points L and N; and let A Land MN be joined. The angle GAL is equal to the angle MDN. Make A H equal to DM, and through H draw HK parallel to GL in the plane AGL, and meeting AL in K. Because GL is perpendicular to the plane BAC, H K is perpendicular to the same plane. Because the solid angle at A, is contained by three plane angles B AC, BAH, HAC, which are equal, each to each, to the three plane angles EDF, EDM, MDF, which contain the solid

પોલી

angle at D. Therefore the solid angles at A and D are equal, and coincide with one another (XI. B); that is, if the plane angle B A C be applied to the plane angle EDF, the straight line A H coincides with DM. Because AH is equal to DM, the point H coincides with the point M. Therefore HK, which is perpendicular to the plane BAC, coincides with MN (XI. 13) which is perpendicular to the plane EDF, and HK is equal to MN. Because the points A and K coincide with the points D and N, the straight line A K coincides with the straight line DN. Therefore the triangle AHK coincides with the triangle DMN, and the angle HAK with the angle MDN. Wherefore the angle G AL is equal to the angle MDN. Therefore, if from the vertices, &c. Q. E. D.

COROLLARY.-From this it is manifest, that if from the vertices of two equal plane angles, there be drawn two equal straight lines containing equal angles with the sides of the angles each to each; the perpendiculars drawn from the extremities of the equal straight lines to the planes of the first angles are equal to one another.

If three straight lines be proportionals, the parallelopiped described from all three, as its sides, is equal to the equilateral parallelopiped described from the mean proportional, one of the solid angles of which is contained by three plane angles equal, each to each, to the three plane angles containing one of the solid angles of the other. Let A, B, C be three proportionals, viz., A to B, as B to C. The