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ibis.

paffing thro' E D, DC, meeting the said Plane in the † 23. 1. 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 E DC, 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' B AC; and make K H equal to GF; and join HA. I say, the solid Angle at A, which is contained under the three plane Angles BAL, BAH, HA L, is equal to the folid Angle at D, which is contained under the plane Angles E DE, EDF, FDC: For let the equal Right Lines A B, D Е, be taken ; and join H B,

KB, FE, GE: Then, becaule F G is perpendicular 1 Def. 3. of to the Plane passing thro’ED, DC, it fhall be I per

pendicular to all the Right Lines touching it, that are in the said Plane: Wherefore both die Angles FGD, FGE, are Right Angles. For the fame Reason, both the Angles HKA, R KB, are Right Angles ; and be. cause the two Sides KA, A B, are equal to the two

Sides GD, DE, each to each, and contain equal An+ 4.1.

gles, the Bale B K shall be † equal to the Bale EG: But K H is also equal o GF; and they contain Right Angles; therefore, HB Inall be + equal to F E. Again, because the two Sides A K, K H, are equal to the two Sides D G G F, and they contain Right Angles ; the Base A H shall be equal to the Base DF: But A B is equal to D E; therefore the two S des 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 78.1, tequal to the Angle EDF: For the fame Reason,

the Angle HAL is equal to the Angle FDC: For, since, it A L 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 B A K, a Part of the one, is put equal to the Angle E D G, a Part of the other; the Angle KAL, remaining, will be equal to the Angle G D C remaining. And because the two Sides K A, A L, are equal to the two Sides GD, DC, and they contain equal Angles; the Base K L will be equal to the Base G C: But K His 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 H'L will be equal to the

Bare

Base FC. Again, because the two Sides HA, AL,
are equal to the two Sides BD, DC; and the Base
HL is equal to the Base FC; the Angle]H A L will
be equal to the Angle FDC: But the Angle B.AL
was made equal to the Angle E DC: Therefore a
solid Angle, is made equal to a felid Angle given ; which
was to be done.
PROPOSITION XXVII.

THEOREM.
Upon Right Line given, to describe a Parallelo-
pipedon,

similar, and in like manner situate, to
a solid Parallelopipédon.

a

Parallelopipedon; ic is required to describe a solid Parallelopipedon upon the given Right Line A B, fimilar, and alike situate, to the given solid Parallelopiper don CD.

Make a folid Angle at the given Point 'A, in the Right Line A B, * contained under the Angle BAH, * 26 of ibis, HÀ K,K AB; fo that the Angle B AH 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 B A + to AK; and + 12. 6. as G C to CF, so K A to AH: Then (by Equality of Proporcion) as E C is to CE, so thall B Abe to AH: Compleat the Parallelogram BH. and the Solid AL; then, because it is, as E C is to GC, so is A B to AK; viz. the Sides about the equal Angles ECG BAK, proportional; the Parallelogram K B Thall be similar to the Parallelogram G E. Also, for the same Reason the Parallelogram ķ H hall be similar to the Parallelogram GF, and the Parallelogram H B to the Parallelogram FE: Therefore three Parallelograms of the Solid A L, are similar to three Parallelograms of the Solid CD. But these three Parallolograms are equal and similar to their three opposite ones; there

1 Cor. 10. fore the whole Solid AL will be similar to the whole of ebis. Solid CD; and so, a solid Parallelopipedon A L is described upon the given Right Line A B, Jimilar and alike fatuate, to the given folid Parallelopipedon CD; which was to be done.

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PROPOSITION XXVIII.

THEOR E M.
If a folid Parallelopipedon be cut by a Plane pal-

fing tbro' tbe Diagonals of two opposite Planes,
ibat Solid will be bifected by the Plane.
ET the folid Parallelopipedon A B be cut by the

Plane CDEF passing thro' the Diagonals CF,
DE, of two opposite Planes. I say, the Solid A B is

I lay, bifected by the Plane C DEF.

For, because the Triangle CGF is * equal to the Triangle CBF, and the Triangle ADE to the Triangle

DEH, and the Parallelogram C A to f the Parallelo+ 24 tbis.

gram B E, for it is opposite 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, 1Def. 10. of

and the three Parallelograms CH, BE, CE; for they are contained under Planes equal in Number and Mag. nitude. Therefore, the whole Solid A B is bifected by the Plane CDEF; which was to be demonstrated,

34. I.

ibis,

PROPOSITION XXIX.

THEOREM.
Solid Parallelopipedons, being constituted upon the

Same Base, and having the same Altitude, and
whose insistent Lines are in the same Right
Lines, are equal to one another.

, conftituted upon the same 'Base A B, with the fame Altitude, whose insistent Lines A F, AG, LM, LN, CD, CE, BH, B K, are in the Same Right Lines FN, DK. I say, the Solid C M is equal to

For, because CH, CK, are both Parallelograms, CB dhall be * equal to DH, or E K; wherefore DH

the Solid B F.

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