† 23. 1.

1

passing thro E D, DC, meeting the said Plane in the
Point G, and join DG; make † the Angles BAL,
BAK, at the given Point A, with the Right Line
A B, equal to the Angles EDC, EDG.

Lastly, make A K equal to DG, and at the Point *12 of this. K erect HK, at Right Angles to the Plane paffing thro' BAL; and make KH equal to GF; and join HA. I say, 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 AB, DE, be taken; and join HB, KB, FE, GE: Then, because FG is perpendicular

* Def. 3. of to the Plane paffing thro' ED, DC, it shall be *

this.

† 4.1.

8. 1.

;

per

pendicular to all the Right Lines touching it, that are in the faid Plane: Wherefore both the Angles FGD, FGE, are Right Angles. For the same Reason, 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 Base BK shall be + equal to the Base EG: But KH is also equal to GF; and they contain Right Angles; therefore, H B shall be + equal to FE. Again, because the two Sides A K, KH, are equal to the two Sides DG, GF, and they contain Right Angles; the Base A H shall be equal to the Base DF: But A B is equal to DE; therefore the two Sides HA, A B, are equal to the two Sides FD, DE. But the Base H B is equal to the Bafe FE; and so the Angle BAH will be I equal to the Angle 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 B A L 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 Base KL will be equal to the Base 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 Base HL will be equal to the Bafe

Base FC. Again, because the two Sides HA, AL, are equal to the two Sides FD, DC, and the Base HL is equal to the Base FC; the Angle HAL will be equal to the Angle FDC: But the Angle BAL was made equal to the Angle EDC: Therefore a

folid Angle is made equal to a folid Angle given; which was to be done.

PROPOSITION XXVII.

THEOREM.

Upon a Right Line given, to describe a Parallelepipedon, fimilar, and in like manner fituate, to a folid Parallepipedon.

LET AB be a Right Line, and CD a given solid Parallelepipedon; it is required to describe a folid Parallepipedon upon the given Right Line A B, fimilar, and alike situate, to the given solid Parallelepipedon CD.

Make a folid Angle at the given Point A, in the Right Line A B, * contained under the Angles BAH, * 26 of thise HAK, KAB; so that the Angle BAH may 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 † 12.6. as GC to CF, fo KA to AH: Then (by Equality of Proportion) as EC is to CF, so shall BA be to AH: Compleat the Parallelogram B H, and the Solid AL; then, because it is, as EC is to GC, so is A B to AK; viz, the Sides about the equal Angles ECG, BAK, proportional; the Parallelogram K B shall be similar to the Parallelogram GE. Also, for the fame Reason, the Parallelogram KH shall be similar to the Parallelogram GF, and the Parallelogram HB to the Parallelogram FE: Therefore three Parallelograms of the Solid A L, are similar to three Parallelograms of the Solid C D. But these three Parallelograms are ‡ equal and fimilar to their three oppofite ones; there-1 Cor. 20. fore the whole Solid AL will be similar to the whole of this. Solid CD; and so, a folid Parallelepipedon AL is deScribed upon the given Right Line AB, fimilar and alike fituate, to the given folid Parallelepipedon CD; which was to be done.

PRO

34. Ι.

PROPOSITION XXVIII.

THEOREM.

1

If a folid Parallelepipedon be cut by a Plane paf-
fing thro' the Diagonals of two opposite Planes,
that Solid will be bisected by the Plane.

LET the folid Parallelepipedon A B be cut by the
Plane CDEF, paffing thro' the Diagonals C F,
DE, of two opposite Planes. I say, the Solid A B is
bisected by the Plane CDEF.

For, because the Triangle CGF is * equal to the Triangle CBF, and the Triangle ADE to the Triangle 24 of this. DEH, and the Parallelogram CA to + the 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 CF B, DEH, and the three Parallelograms CH, BE, CE; for ‡ they are contained under Planes equal in Number and Magnitude. Therefore, the whole Solid A B is bisected by the Plane CDEF; which was to be demonftrated.

Def. 10.

of this.

PROPOSITION ΧΧΙΧ.

THEOREM.

Solid Parallelepipedons, being constituted upon the fame Bafe, and having the fame Altitude, and whose insistent Lines are in the same Right Lines, are equal to one another.

LET the folid Parallelepipedons CM, BF, be constituted upon the same Bafe A B, with the same Altitude, whose insistent Lines AF, AG, LM, LN, CD, CE, BH, BK, are in the same Right Lines FN, DK. I fay, the Solid CM is equal to the Solid B F.

For, because CH, CK, are both Parallelograms, CB shall be * equal to DH, or EK; wherefore DH