+23.1.

tbis.

† 4.1.

paffing thro' ED, DC, meeting the faid Plane in the
Point G, and join DG; make + the Angles B A L,
BAK, at the given Point A, with the Right Line
AB, equal to the Angles EDC, E DG.

Laftly, 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 A C; and make K H equal to GF; and join HA. I fay, the folid Angle at A, which is contained under the three plane Angles B A L, 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, D E, be taken; and join H B, KB, FE, GE: Then, becaufe F G is perpendicular Def. 3. of to the Plane paffing thro' ED, DC, it fhall be perpendicular to all the Right Lines touching it, that are in the faid Plane: Wherefore both the Angles FGD, FGE, are Right Angles. For the fame Reason, both the Angles H KA, HK B, are Right Angles; and because the two Sides K A, A B, are equal to the two Sides GD, D E, each to each, and contain equal Angles, the Bafe BK fhall be equal to the Bafe EG: But K H is alfo equal to GF; and they contain Right Angles; therefore, H B fhall be † equal to F E. Again, because the two Sides A K, K H, are equal to the two Sides DG, GF, and they contain Right Angles; the Bafe A H fhall be equal to the Bafe DF: But A B is equal to DE; therefore the two Sides HA, A B, are equal to the two Sides F D, D E. But the Bafe HB is equal to the Bafe F E; and fo the Angle BA H will be + equal to the Angle EDF: For the fame Reafon, the Angle HAL is equal to the Angle FDC: For, fince, if A L be taken equal to DC; and K L, HL, GC, FC, be joined; the whole Angle B A L is equal to the whole Angle EDC; and the Angle BA K, a Part of the one, is put equal to the Angle E DG, a Part of the other; the Angle K A L, 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 KL will be equal to the Bafe GC: But K His equal to GF; wherefore the two Sides L K, KH, are equal to the two Sides CG, GF: But they contain Right Angles; therefore the Bafe HL will be equal to the

Ba

Bafe

a C. Again, because the two Sides HA, AL,

are equal to the two Sides B D, DC; and the Bafe HL is equal to the Bafe FC; the Angle H A L will be equal to the Angle F DC: But the Angle BAL was made equal to the Angle EDC: Therefore a Solid Angle, is made equal to a folid Angle given; which was to be done.

PROPOSITION XXVII

THEOREM.

Upon a Right Line given, to defcribe a Parallelopipedon, fimilar, and in like manner fituate, to a folid Parallelopipedon.

LET AB be a Right Line, and CD a given folid Parallelopipedon; it is required to defcribe a folid Parallelopipedon upon the given Right Line A B, fimilar, and alike fituate, to the given folid Parallelopipe don CD.

Make a folid Angle at the given Point A, in the Right Line A B, contained under the Angle B A H, * 26 of this, HAK,KAB; fo that the Angle B AH may be equal to the Angle ECF, the Angle BAK to the Angle ECG, and the Angle H A K to the Angle GCF; and make, as EC is to CG, fo BA+ to AK; and + 12. 6. as GC to CF, fo KA to AH: Then (by Equality of Proportion) as EC is to CF, fo fhall B A be to AH: Compleat the Parallelogram B H, and the Solid AL; then, because it is, as EC is to GC, fo is A B to A K; viz. the Sides about the equal Angles ECG BAK, proportional; the Parallelogram KB fhall be fimilar to the Parallelogram G E. Alfo, for the fame Reason the Parallelogram KH fhall be fimilar to the Parallelogram GF, and the Parallelogram H B to the Parallelogram FE: Therefore three Parallelograms of the Solid A L, are fimilar to three Parallelograms of the Solid CD. But these three Parallolograms are equal and fimilar to their three oppofite ones; there↑ Cor. 20. fore the whole Solid A L will be fimilar to the whole of this. Solid CD; and fo, a folid Parallelopipedon A L is defcribed upon the given Right Line A B, fimilar and alike fituate, to the given folid Parallelopipedon CD; which was to be done.

PRO

† 24 of this.

IDef. 10.

this.

34.

PROPOSITION XXVIII.

THEOREM.

If a folid Parallelopipedon 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 Parallelopipedon A B be cut by the Plane C D E F paffing thro' the Diagonals CF, DE, of two oppofite Planes. I fay, the Solid A B is bifected by the Plane C DE F.

For, because the Triangle C G F is equal to the Triangle CBF, and the Triangle ADE to the Triangle DEH, and the Parallelogram C A to + the Parallelogram B E, for it is oppofite to it: and the Parallelogram G E to the Parallelogram CH; the Prifm contained by the two Triangles CGF, ADE, and the three Parallelograms GE, AC, CE, is equal to the Prifm contained under the two Triangles CFB, DEH, and the three Parallelograms CH, BE, CE; for t they of 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 demonftrated,

PROPOSITION XXIX.

THEOREM.

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

LET the folid Parallelopipedons CM, BF, be

conftituted upon the fame Bafe A B, with the fame Altitude, whofe infiftent Lines A F, 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 B F.

For, becaufe C H, CK, are both Parallelograms, CB fhall be equal to DH, or E K; wherefore D H