this.

lelograms HG, HI, IN; and fo are the Parallelograms DH, M2, NT. Therefore three Planes of the Solid L P, are equal to three Planes of the Solid KR, or AY, each to each; and the Planes oppofite to these, are equal to them. Therefore the three Solids 1 Def. 10. of LP, KR, ÂY, will be equal ‡ to each other. For the same Reason, the three Solids ED, Ha, MT, are equal to each other. Therefore the Bafe LF is the fame Multiple of the Bafe AF, as the Solid LY is of the Solid AY. For the fame Reason, the Base NF is the fame Multiple of the Base HF, as the Solid NY is of the Solid ED: And if the Base LF be equal to the Bafe NF, the Solid LY fhall be equal to the Solid NY; and if the Base LF exceeds the Base NF, the Solid LY fhall exceed the Solid NY; and if it be lefs, lefs. Wherefore because there are four Magnitudes, viz. the two Bases AF, FH, and the two Solids AY, ED, whose Equimultiples are taken, to wit, the Bafe LF, and the Solid LY; and the Base NF, and the Solid NY: And fince it is proved, if the Base LF exceeds the Base NF, then the Solid LY will exceed the Solid NY, if equal, equal, and less, less. Therefore as the Base AF is Def. 6. 5. to the Base F H, fo is * the Solid AY to the Solid ED. Wherefore, if a folid Parallelepipedon be cut by a Plane, parallel to oppofite Planes; then as Bafe is to Bafe, fo fhall Solid be to Solid; which was to be demonftrated.

PROPOSITION XXVI.

THEOREM.

At a Right Line given, and at a Point given in it, to make a folid Angle equal to a folid Angle given.

ET AB be a Right Line given; A a given Point in it, and D a given folid Angle contained under the plane Angles EDC, EDF, FDC; it is required to make a folid Angle at the given Point A, in the given Right Line AB, equal to the given folid Angle D.

Affume any Point F in the Right Line DF, from * 11 of this. which let FG be drawn * perpendicular to the Plane paffing

paffing thro' ED, DC, meeting the said Plane in the Point G, and join DG, make† the Angles BAL, † 23. I. BAK, at the given Point A, with the Right Line AB, equal to the Angles EDC, EDG.

Laftly, make AK equal to DG, and at the Point

this.

K erect HK at Right Angles to the Plane paffing ‡ 12 of this thro' BAL, and make KH equal to GF, and join HA. I fay, the solid Angle at A, which is contained under the three plane Angles BAL, BAH, HAL, is equal to the folid Angle at D, which is contained under the plane Angles EDC, EDF, FDC: For/ let the equal Right Lines A B, DE, be taken, and join HB, KB, FE, GE. Then because F G is perpendicular to the Plane paffing through ED, DC, it fhall be perpendicular to all the Right Lines touch-* Def. 3.f ing it that are in the faid Plane. Wherefore both the Angles FGD, FGE, are Right Angles. For the fame Reafon, both the Angles HKA, HKB, are Right Angles; and because the two Sides KA, AB, are equal to the two Sides GD, DE, each to each, and contain equal Angles, the Bafe BK fhall be equal to the † 4. 14 Bafe EG; but KH is alfo equal to GF, and they contain Right Angles; therefore HB fhall be + equal to FE. Again, because the two Sides AK, KH, are equal to the two Sides DG, GF, and they contain Right Angles, the Base AH fhall be equal to the Base DF; but AB is equal to DE. Therefore the two Sides HA, A B, are equal to the two Sides F D, DE; but the Base H B is equal to the Base F E, and fo the Angle BAH will be equal to the Angle‡ 3. L EDF. For the fame Reafon, the Angle HAL is equal to the Angle FDC; for fince if AL be taken equal to DC, and KL, HL, GC, FC, be joined, the whole Angle BAL is equal to the whole Angle EDC; and the Angle BAK, a Part of the one, is put equal to the Angle EDG, a Part of the other; the Angle KAL remaining, will be equal to the Angle GDC remaining. And because the two Sides KA, AL, are equal to the two Sides GD, DC, and they contain equal Angles, the Bafe KL will be equal to the Bafe GC; but KH is equal to GF; wherefore the two Sides LK, KH, are equal to the two Sides CG, GF, but they contain Right Angles; therefore the Bafe HL will be equal to the Baie FC. P 4

Again,

† 12. 6.

Again, because the two Sides HA, AL, are equal to the two Sides F D, DC, and the Bafe HL is equal to the Bafe FC, the Angle HAL will be equal to the Angle FDC; but the Angle BAL is equal to the Angle EDC; which was to be done.

PROPOSITION XXVII.

PROBLEM.

Upon a Right Line given, to defcribe a Parallelepipedon fimilar, and in like manner situate to a Jolid Parallelepipedon given.

LEParallelepipedon.

ET AB be a Right Line, and CD a given folid Parallelepipedon. It is required to defcribe a for lid Parallelepipedon upon the given Right Line AB, fimilar and alike fituate to the given folid Parallelepipedon CD.

so

Make a folid Angle at the given Point A, in the 26 of this. Right Line A B, which is contained under the Angles BAH, HAK, KAB; fo that the Angle BAH be equal to the Angle ECF, the Angle BAK to the Angle ECG, and the Angle HAK to the Angle GCF; and make as EC is to CG, so BA † to AK; and GC to CF, as KA to AH. Then (by Equality of Proportion) as EC is to CF, fo fhall BA be to AH; complete the Parallelogram BH, and the Solid A L. Then because it is as EC is to CG, fo is BA to AK, viz. the Sides about the equal Angles ECG, BAK, proportional; the Parallelogram KB, fhall be fimilar to the Parallelogram GE. Alfo, for the fame Reason, the Parallelogram KH, fhall be fimilar to the Parallelogram GF, and the Parallelo¬ gram HB, to the Parallelogram FE. Therefore three Parallelograms of the folid A L, are fimilar to three Parallelograms of the Solid CD; but thefe three Parallelograms are equal and fimilar to their three oppofite ones. Therefore the whole Solid AL, will be fimilar to the whole Solid CD; and fo ́a folid Parallelepipedon AL, is defcribed upon the given Right Line A B fimilar, and alike fituate to the given folid Parallelepipedon CD; which was to be done.

+Cor. 24. of this.

PRQ

PROPOSITION XXVIII.

THEOREM.

If a folid Parallelepipedon be cut by a Plane paffing thro' the Diagonals of two oppofite Planes, that Solid will be bifected by the Plane.

L

ET the folid Parallelepipedon AB, be cut by the Plane CDEF, paffing thro' the Diagonals CF, DE, of two oppofite Planes. I fay, the Solid A B is bifected by the Plane CDEF.

For because the Triangle CGF is equal to the* 34. 1. Triangle CBF, and the Triangle ADE to the Triangle DEH, and the Parallelogram CA to † the‍ 24 of this. Parallelogram BE, for it is oppofite to it; and the Parallelogram GE to the Parallelogram CH; the Prism contained by the two Triangles CGF, ADE, and the three Parallelograms GE, AC, CE, is equal to the Prism contained under the two Triangles CFB, DEH, and the three Parallelograms CH, BE, CE; for they are contained under Planes equal in Number and Magnitude. Therefore the whole Solid AB is bifected by the Plane CDEF; which was to be demonftrated.

PROPOSITION XXIX.

Solid Parallelepipedons, being conftituted upon the fame Bafe, and having the fame Altitude, and whofe infiftent Lines are in the fame Right Lines, are equal to one another.

ET the folid Parallelepipedons CM, CN, be
conftituted upon

fame Altitude, whofe infiftent Lines AF, AG, LM,
LN, CD, CE, BH, BK, are in the fame Right
Lines FN, DK. I fay, the Solid CM is equal to
the Solid CN.

For becaufe CH, CK, are both Parallelograms, CB fhall be equal to DH, or EK; wherefore DH * 34 L

is

+ S. 1.

is equal to EK. Let EH, which is common, be taken away, then the Remainder DE will be equal to the Remainder HK; and fo the Triangle DEC is +equal to the Triangle HKB, and the Parallelogram DG equal to the Parallelogram HN. For the fame Reafon, the Triangle AFG is equal to the Triangle 124 of this. LMN; and the Parallelogram CF to the Parallelogram BM, and the Parallelogram CG to the Parallelogram BN, for they are oppofite. Therefore the Prism contained under the two Triangles AFG, DEC, and the three Parallelograms AD, DG, GC, is equal to the Prifm contained under the two Triangles LMN, HBK, and the three Parallelograms BM, NH, BN. Let the common Solid, whofe Bafe is the Parallelogram AB, oppofite to the Parallelogram GEHM, be added; then the whole folid Parallelepipedon CM, is equal to the whole folid Parallelepipedon CN. Therefore, folid Parallelepipedons, being conftituted upon the fame Bafe, and having the fame Altitude, and whose infiftent Lines are in the fame Right Lines, are equal to one another; which was to be demonstrated.

PROPOSITION XXX.

THEOREM.

Solid Parallelepipedons, being conftituted upon the fame Bafe, and having the fame Altitude, whofe infiftent Lines are not placed in the fame Right Lines, are equal to one another.

LET there be folid Parallelepipedons CM, CN,

having equal Altitudes, and standing on the fame Bafe AB,. and whofe infiftent Lines AF, AG, LM, LN, CD, CE, BH, BK, are not in the fame Right Lines. I fay, the Solid CM is equal to the Solid CN.

For let NK, DH, and GE, FM, be produced, meeting each other in the Points R, X; let alfo FM, GE, be produced to the Points O,.P, and join AX, LO, CP, BR. The Solid CM, whofe Bafe is the Parallelogram ACBL, being oppofite to the Paral*29 of this. lelogram FDHM, is equal to the Solid CO, whofe

*