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solid AB to the solid EX, 80 is EX the solid CD, and the straight line to PL, and PL to KO: But if four EK is equal to the straight line CF. magnitudes be continual proportion. Therefore the solid AB has to the soals, the first is said to have to the lid CD, the triplicate ratio of that fourth the triplicate ratio of that which which the side AE has to the homoit has to the second : Therefore the logous side CF, &c. Q. E. D. solid AB has to the solid KO, the tri- Cor. From this it is manifest, that, plicate ratio of that which AB has to if four straight lines be continual proEX: But as AB is to EX, so is the portionals, as the first is to the fourth, parallelogram AG to the parallelo- so is the solid parallelopiped described gram GK, and the straight line AE from the first to the similar solid simito the straight line EK. Wherefore larly described from the second; bethe solid AB has to the solid KO, the cause the first straight line has to the triplicate ratio of that which AE bas fourth the triplicate ratio of that which to EK. And the solid KO is equal to it has to the second.

PROP. D. THEOREM.

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Solid parallelopipeds contaimed 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 AB, CD be solid parallelopi- Book 6, and the solids AB, AY, being peds, of which AB is contained by the betwixt the parallel planes BOY, parallelograms AE, AF, AG equian- EAS, are of the same altitude. Theregular, each to each, to the parallelo-fore the solid AB is to the solid AY, grams CH, CK, CL, which contain

as (32. 11.) the base AE to the base the solid CD. The ratio which the AS; that is, as the straight line a is solid AB has to the solid CD, is the to c. And the solid AY is to the sosame with that which is compounded of the ratios of the sides AM to DL, AN to DK, and AO to DH.

Produce MA, NA, OA to P, Q, R, so that AP be equal to DL, AQ to DK, and AR to DH; and complete the solid parallelopiped AX contained

м by the parallelograms AS, AT, AV similar and equal to CH, CK, CL, each to each. Therefore the solid AX is equal (C. 11.) to the solid CD. Complete likewise the solid AY, the base of which is AS, and of which AO lid AX, as (25. 11.) the base OQ is to is one of its insisting straight lines. the base QÅ; that is, as the straight Take any straight line a, and as MA Line OA to AR: that is, as the straight to AP, so make a to b; and as NA line c to the straight line d. And beto AQ, so make b to c; and as AO to cause the solid AB is to the solid AY, AR, so c to d: Then, because the pa- as a is to c, and the solid AY to the rallelogram AE is equiangular to ås, solid AX, as c is to d; ex æquali, the AE is to AS, as the straight line a tó solid AB is to the solid AX, or CD c, as is demonstrated in the 23d Prop. which is equal to it, as the straight

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line a is to d. But the ratio of a to equal to the sides DL, DK, DH, each d is said to be compounded (def. A. to each. Therefore the solid AB has 5.) of the ratius of a to b, b to c, to the solid CD the ratio, which is the and c to d, which are the same with same with that which is compounded the ratios of the sides MA to AP, of the ratios of the sides AM to DL, NA to AQ, and OA to AR, each to AN to DK, and AO to DH, Q. E. D. each. And the sides AP, AQ, AR are

PROB. XXXIV. THEOREM. The bases and altitudes of equal solid parallelopipeds, are reciprocally

proportioned; and if the bases and altitudes be reciprocally proportional, the solid parallelopipeds are equal. Let AB, CD be equal solid paral- other: Since then the solid AB is elelopipeds; :heir bases are reciprocally qual to the solid CD, CM is therefore proportional to their altitudes : that greater than AG: For if it be not, is, as the base EH is to the base NP, neither also in this case would the soso is the altitude of the solid CD to lids AB, CD be equal, which, by the the altitude of the solid AB.

hypothesis, are equal. Make then CT First, Let the insisting straight lines equal to AG, and complete the solid AG, EF, LB, HK; CM, NX, OD, parallelopiped CV of which the base PR be at right angles to the bases. is NP, and altitude CT. Because the As the base EH to the base NP, so is solid AB is equal to the solid CD, CM to AG. If the base EH be equal therefore the solid AB is to the solid to the base NP, then because the so- CV, as (7.5.) the solid CD to the solid AB is likewise equal to the solid lid CV. But as the solid AB to the CD, CM shall

solid CV, so (34. 11.) is the base EH K be equal to

to the base NP; for the solids AB, AG. Because

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X CV are of the same altitude; and as if the bases

the solid CD to CV, so (25. 11.) is the H EH, NP, be

base MP to the base PT; and so (1. A equal, but the

6.) is the straight line MC to CT; and altitudes AG, CM be not equal, nei. CT is equal to AG. Therefore, as the ther shall the solid AB be equal to the base EH to the base NP, so is MC to solid CD. But the solids are equal, AG. Wherefore the bases of the soby the hypothesis. Therefore the als lid parallelopipeds AB, CD are retitude CM is not unequal to the alti- ciprocally proportional to their altitude AG; that is, they are equal. tudes. Wherefore as the base EH to the base . Let now the bases of the solid paNP, so is CM to AG.

rallelopipeds AB, CD be reciprocally Next, Let the bases EH, NP not proportional to their altitudes; viz. be equal, but EH greater than the as the base EH to the base NP, 80

the altitude of the solid CD to the al

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titude of the solid AB; the solid AB D, R, M draw perpendiculars to the is equal to the solid CD. Let the in- planes in which are the bases EH, sisting lines be, as before, at right an- NP meeting those planes in the points gles to the bases. Then, if the base S, Y, V, T; Q, I, U, Z; and comEH be equal to the base NP, since plete the solids FÝ, XU, which are EH is to NP, as the altitude of the parallelopipeds, as was proved in the solid CD is to the altitude of the solid last part of Prop. XXXI. of this AB, therefore the altitude of CD is e. Book. In this case, likewise, if the qual (A. 5.) to the altitude of AB. solids AB, CD be equal, their bases But solid parallelopipeds upon equal are reciprocally proportional to their bases, and of the same altitude, are altitudes, viz. the base EH to the base equal (31. 11.) to one another; there. NP, as the altitude of the solid CD to fore the solid AB is equal to the solid the altitude of the solid AB. Because CD.

the solid AB'is equal to the solid CD, But let the bases EH, NP be un, and that the solid BT is equal (29. or equal, and let EH be the greater of 30. 11.) to the solid BA, for they are the two. Therefore, since as the base upon the same base FK, and of the EH to the base NP, so is CM the al

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titude of the solid CD to AG the alti- game altitude; and that the solid DC tude of AB, CM is greater (A. 5.) than is equal (29. or 30. 11.) to the solid AG. Again, take CT equal to AG, DZ, being upon the same base XR, and complete, as before, the solid CV. and of the same altitude; therefore And, because the base EH is to the the solid BT is equal to the solid base NP, as CM to AG, and that AG DZ: But the bases are reciprocally is equal to CT, therefore the base proportional to the altitudes of eEH is to the base NP, as MC to CT. qual solid parallelopipeds of which But as the base EH is to NP, 20 (32. the insisting straight lines are at 11.) is the solid AB to the solid CV; right angles to their bases, as befor the solids AB, CV are of the same fore was proved : Therefore as the altitude: and as MC to C'T, so is the base FK to the base XR, so is the base MP to the base PT, and the 80- altitude of the solid DZ to the altitude lid CD to the solid (25.11.) CV; And of the solid BT: And the base FK is therefore as the solid AB to the solid equal to the base EH, and the base CV, so is the solid CD to the solid XR to the base NP: Wherefore, as CV; that is, each of the solids AB, the base EH to the base NP, so is the CD has the same ratio to the solid CV; altitude of the solid DZ to the altitude and therefore the solid AB is equal to of the solid BT: But the altitudes of the solid CD.

the solids DZ, DC, as also of the sou Second general case.

Let the in- lids BT, BA are the sarne. Therefore sisting straight lines FE, BL, GA, as the base EH to the base NP, so is KH; XN, DO, MC, RP not be at the altitude of the solid CD to the alright angles to the bases of the solide; titude of the solid AB; that is, the and from the points F, B, K, G; X, bases of the solid parallelopipeds AB, CD are reciprocally proportional to base FK; and NP to XR; therefore their altitudes.

the base FK is to the base XR, as the Next, Let the bases of the solids altitude of the solid CD to the altitude AB, CD be reciprocally proportional of AB: But the altitudes of the solids to their altitudes, viz. the base EH to AB, BT are the same, as also of CD the base NP, as the altitude of the and DZ; therefore as the base FK to solid CD to the altitude of the solid the base XR, so is the altitude of the AB; the solid AB is equal to the so- solid DZ to the altitude of the solid lid CD: The same construction being BT: Wherefore the bases of the somade; because as the base EH to the lids BT, DZ are reciprocally proporbase NP, so is the altitude of the so- tional to their altitudes; and their inlid CD to the altitude of the solid AB; sisting straight lines are at right anand that the base EH is equal to the gles to the bases; wherefore, as was

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before proved, the solid BT is equal K B

to the solid DZ: But BT is equal X

(29. or 30. 11.) to the solid BA; and DZ to the solid DC, because they are

upon the same bases, and of the same II WIIL 'SP

altitude. Therefore the solid AB is equal to the solid CD. Q. E. D.

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PROP. XXXV. THEOREM.

If, from the vertices of two equal plane angles, there be drawn tro

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 the planes in which the first named angles are: And from the points in which they meet the planes, straight lines be drawn to the vertices of the angles first named; these straight lines shall contain equal angles with the straight lines which are above the planes of the angles.

Let BAC, EDF be two equal plane culars GL, MN be drawn to the planes angles; and from the points A, D let BAC, EDF, meeting these planes in the straight lines AG, DM be elevated the points L, N; and join LA, ND: above the planes of the angles, mak. The angle GAL is equal to the angle ing equal angles with their sides, each MDN. to each, viz. the angle GAB equal to Make AH equal to DM, and through the angle MDE, and GAC to MDF; H draw HK parallel to GL: But GL and in AG, DM let any points G, M is perpendicular to the plane BAC; be taken, and from them let perpendi- wherefore HK is perpendicular (8. 11.)

to the same plane: From the points A D

K, N, to the straight lines AB, AC,

DE, DF, draw perpendiculars KB, B

KC, NE, NF; and join HB, BC, ME,

EF: Because HK is perpendicular to M

the plane BAC, the plane HBK which passes through HK is at right angles

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(18. 11.) to the plane BAC; and AB gle BCK is equal to the angle EFN: is drawn in the plane BAC at right Therefore in the two triangles BCK, angles to the common section BK of EFN, there are two angles in one e the two planes: Therefore AB is per- qual to two angles in the other, each pendicular (4. def. 11.) to the plane to each, and one side equal to one HBK, and makes right angles (3. def. side adjacent to the equal angles in 11.) with every straight line meeting each, viz. BC equal to EF; the other it in that plane : But BH meets it in sides, therefore, are equal to the other that plane; therefore ABH is a right sides; BK then is equal to EN: And angle: For the same reason, DEM is a AB is equal to DE; wherefore AB, right angle, and is therefore equal to BK are equal to DE, EN; and they the angle ABH: And the angle HAB contain right angles ; wherefore the is equal to the angle MDE. There- base AK is equal to the base DN: fore in the two triangles HAB, MDE And since AH is equal to DM, the there are two angles in one equal to square of AH is equal to the square of two angles in the other, each to each, DM: But the squares of AK, KH are and one side equal to one side, oppo- equal to the square (47. 1.) of AH, site to one of the equal angles in each, because AKH is a right angle: And viz. HA equal to ŪM; therefore the the squares of DN, NM are equal to remaining sides are equal, (26. 1.) the square of DM, for DNM is a right each to each: Wherefore AB is equal angle: Wherefore the squares of AK, to DE. In the same manner, if HC KI are equal to the squares of DN, and MF be joined, it may be demon- NM; and of those the square of AK strated that AC is equal to DF: is equal to the square of DN: ThereTherefore, since AB is equal to DE, fore the remaining square of KH is BA and AC are equal to ÉD and DF; equal to the remaining square of NM;

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and the straight line KH to the straight line NM: And because HA, AK are

equal to MD, DN, each to each, ani B

the base HK to the base MN, as has L E 팬

been proved; therefore the angle HAK M

is equal (8.1.) to the angle MDN. and the angle BAC is equal to the an, Q.E.D. gle EDF; wherefore the base BC is Cor. From this it is manifest, that equal (1. 1.) to the base EF, and the if, from the vertices of two equal plane remaining angles to the remaining an- angles, there be elevated two equal gles : The angle ABC is therefore ea straight lines containing equal angles qual to the angle DEF: And the right with the sides of the angles, each to angle ABK is equal to the right angle each; the perpendiculars drawn from DEN, whence the remaining angle the extremities of the equal straght CBK is equal to the remaining angle lines to the planes of the first angles FEN: For the same reason, the an

are equal to one another.

Another Demonstration of the Corollary. Let the plane angles BAC, EDF be qual to MDE, and HAC equal to the equal to one another, and let AH, angle MDF; and from H, M let HK, DM be two equal straight lines above MN be perpendiculars to the planes the planes of the angles, containing e- BAC, EDF: HK is equal to MN. qual angles with BA, AC; ED, VF, Because the solid angle at A is coneach to each, viz. the angle HAB, e- tained by the three plane angles BAC

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