*

XM, and XR is common, and at Right Angles to them, the Bafe LR fhall be equal to the Bafe R M, * 4« 1. For the fame Reafon, R N is equal to R E, or RM; therefore three Right Lines R L, R M, RN, are equal to each other. And because the Square of X R is equal to the Excess by which the Square of A B exceeds the Square of L X, the Square of A B will be equal to the Squares of LX, X R, together: But the Square of R L is † equal to the Squares of L X, XR for L XR is at 47. 1. Right Angle; therefore the Square of A B will be equal to the Square of RL; and fo A B is equal to R L. But B C, DE, EF, GH, HK, are every one of them. equal to A B; and R N or R M, equal to RL; wherefore A B, BC, DE, EF, G H, HK, are each equal to R L, R M, or RN: And fince the two Sides RL, RM, are equal to the two Sides A B, B C ; and the Bafe L M is put equal to the Base A C; the Angle L R M fhall be equal to the Angle ABC. For 1 s. 1, the fame Reason the Angle M R N is equal to the Angle DEF, and the Angle LR N equal to the Angle GHK: Therefore, a folid Angle is made at R of three plane Angles LRM, MRN, LRN, equal to three plane Angles given ABC, DEF, GHK.

Now, let the Centre of the Circle X be in one Side of the Triangle, viz. in the Side M N; and join X L. I fay, again, that A B is greater than LX. For, if it be not fo, A B will be either equal, or lefs than L X. Firft, let it be equal; then the two Sides A B, B C, are equal to the two Sides M X, L X, that is, they are equal to MN: But M N is put equal to D F ; there fore DE, EF, are equal to D F, which is impoffible; * 20. ■ therefore A B is not equal to L X. In like manner we prove, that it is neither leffer; for the Abfurdity will much more evidently follow. Therefore A B is greater than LX. And if the Square of R X be made equal to the Excess by which the Square of A B exceeds the Square of L X, and R X be raised at Right Angles to the Plane of the Circle, the Problem may be done in like manner as before.

Laftly, Let the Centre X of the Circle be without the Triangle L M N, and join L X, MX, NX: I fay, A B is greater than LX. For, if it be not, it muft either be equal, or lefs. First, let it be equa':

+ 8.1.

2. 6.

+ 4.6.

then the two Sides AB, BC, are equal to the two Sides M X, X L, each to each; and the Base AC is equal to the Base M L; therefore the Angle A B C is + equal to the Angle MXL. For the fame Reafon, the Angle G HK is equal to the Angle LX N; and fo the whole Angle MXN is equal to the two Angles ABC, GHK: But the Angles A BC, GHK, are greater than the Angle DEF; therefore the Angle MXN is greater than DEF; But because the two Sides DE, E F, are equal to the two Sides M X, X N, and the Bafe D F is equal to the Base MN: the Angle MN X fhall be equal to the Angle DEF: But it has been proved greater, which is abfurd; therefore AB is not equal to LX. Moreover, we will prove, that it is not lefs; wherefore it shall be neceffarily greater. And if, again, X R be raised at Right Angles to the Plane of the Circle, and made equal to the Side of that Square by which the Square of A B exceeds the Square of LX; the Problem will be determined. Now, I fay, A B is not less than LX: For, if it is poffible that it can be lefs, make XO equal to AB, and X P equal to B C, and join OP; then because A B is equal to B C, XO fhall be equal to X P, and the remaining Part OL equal to the remaining Part P M; therefore L M is parallel to PO, and the Triangle LMX equiangular to the Triangle PXO: Wherefore, as + XL is to LM, fo is XO to OP; and (by Alternation) as LX is to X O, fo is L M to OP: But LX is greater than X O; therefore L M is greater than OP; but L M is equal to A C; wherefore A C fhall be greater than O P: And fo, because the two Sides A B, BC, are equal to the two Sides OX, XP, each to each; and the Bafe A C is greater than the Base OP; the Angle ABC fhall be greater than the Angle OX P. So, likewife, if X R be taken equal to X 0, or X P, and OR be joined, we prove, that the Angle GHK is greater than the Angle OX R. At the Point X, with the Right Line L X, make the Angle L X S equal to the Angle ABC, and the Angle LXT equal to the Angle GHK, and XS, XT, each equal to XO, and join OS, OT, ST; then, becaufe the two Sides A B, BC, are equal to the two Sides O X, X S, and the Angle ABC is equal to the Angle OXS, the Bafe

A C.

A C, that is, LM fhall be equal to the_Bafe O $. For the fame Reafon, L N is alfo equal to OT: And fince the two Sides M L, LN, are equal to the two Sides OS, OT; and the Angle MLN, or POR, is evidently greater than the Angle SOT; the Base MN fhall be greater than the Bafe ST. But MN is equal to DF; therefore D F fhall be greater than ST. Wherefore, becaufe the two Sides EF, DE, are equal to the two Sides S X, XT; and the Base DF is greater than the Bafe ST; the Angle DEF fhall be greater than the Angle S XT. But the Angle SXT is equal to the Angles ABC, GHK ; therefore the Angle DEF is greater than the Angles ABC, GHK: But it is alfo lefs (by Hyp.) which is abfurd; and confequently, A B is not lefs than L X. And fo, a folid Angle may be made of three plane Angles that have the neceffary Limitations; which was to be done.

PROPOSITION

THE ORE M.

XXIV.

If a Solid be contained under fix parallel Planes,
the oppofite Planes thereof are equal Parallelo-
grams.

L
ET the Solid CD GH be contained under paral-
lel Planes A C, GF, BG, CE, F B, A E. I fay,
the oppofite Planes thereof are equal Parallelograms.
For, because the parallel Planes BG, CE, are cut
by the Plane A C, their common Sections *
are paral-
lel; wherefore A B is parallel to CD. Again, becaufe
the two parallel Planes BF, A E, are cut by the Plane
AC, their common Sections are parallel; therefore
AD is parallel to BC. But A B has been proved to
be parallel to C D; wherefore A e fhall be a Parallelo-
gram. After the fame manner we demonftrate, that

16 of this.

E, FG, GB, B F, and A E, are Parallelograms. Let AH, DF, be joined: Then, because A B is parallel to DC, and BH to CF; the Lines AB, BH, touching each other, fhall be parallel to the Lines DC, CF, touching each other, and not being in the fame Plane; wherefore they fhall + contain equal An- †10cfibis. gles: And fo the Angle AB H is equal to the Angle DCF.

P 4

*

DCF. And finee the two Sides A B, BH, are equal to the two Sides DC, CF, and the Angle ABH equal to the Angle DCF; the Bafe AH fhall be equal to the Bafe D F, and the Triangle A B H equal to the Triangle DFC: And fince the Parallelogram BG is + double to the Triangle A B H, and the Parallelogram C E to the Triangle DCF; the Parallelogram BG fhall be equal to the Parallelogram C E, In like manner we demonftrate, that the Parallelogram A C is equal to the Parallelogram G F, and the Parallelogram AE equal to the Parallelogram B F. If, therefore, a Solid be contained under fix parallel Planes, the oppofite Planes thereof are Parallelograms; which was to be demonstrated,

Coroll. It follows, from what has been now demon

ftrated, that, if a Solid be contained under fix paral· Jel Planes, the oppofite Planes thereof are fimilar and equal, because each of the Angles are equal, and the Sides about the equal Angles are proportional.

PROPOSITION XXV.

THEOREM.

If a foid Parallelopipedon be cut by a Plane parallel to oppofite Planes, then, as Base is to Bafe, fo is Solid to Solid.

LET the folid Parallelopipedon ABCD be cut by

a Plane Y E, parallel to the opposite Plane R A, DH. I fay, as the Bafe E FA is to the Base EHCF, fo is the Solid ABFY to the Solid EGCD.

For, let AH be both ways produced, and make HM, MN, &c. equal to EH, and A K, KL, &c. equal to A E; and let the Parallelograms L O, K 0, HX, MS, as likewife the Solids L P, KR, HN, MT, be compleated: Then because the Right Lines LK, KA, AE, are equal; the Parallelograms LO, KO, A F, fhall alfo be equal; as likewife the Parallelograms KE, K B, AG: And moreover, † the Parallelograms LY, K P, AR, for they are oppofite to one another. For the fame Reasons, the Parallelograms EC, HX, MS, alfu, are equal to each other; as alfo the Paral

*

*

Parallelograms HG, HI, IN; and fo are the Parallelograms DH, MO, NT: Therefore three Planes of the Solid L P are equal to three Planes of the Solid KR, or A Y, each to each; and the Planes oppofite to these are equal to them: Therefore the three Solids LP, KR, RY, will be equal to each other. For the fame Reason, the three Solids ED, HQ, MT, are equal to each other; therefore the Bafe L F is of the Multiple of the Base A F, as the Solid LY is of the Solid A Y. For the fame Reason, the Bafe NF is the fame Multiple of the Bafe H F, as the folid NY is of the Solid ED; and if the Base L F be equal to the Bafe NF, the Solid LY fhall be equal to the Solid NY; and if the Bafe L F exceeds the Bafe N F, the Solid LY fhall exceed the Solid NY; and if it be lefs, lefs; Wherefore, because there are four Magnitudes, viz. the two Bafes A F, F H, and the two Solids A Y, ED, whofe Equimultiples are taken, to wit, the Bafe L F, and the Solid LY; and the Bafe NF, and the Solid NY: And fince it is proved, if the Base LF exceeds the Bafe N F, then the Solid LY will exceed the Solid N Y; if equal, equal; and if lefs, less: Therefore as the Base AF is to the Base FH,

*

*

of

Def. 10

this.

fo is the Solid A Y to the Solid E D. Wherefore, Def. 5.5% if a folid Parallelopipedon 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.

PROBLEM.

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

LET AB be a Right Line given, A a given Point in

it, and Da 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 A B, equal to the given folid Angle D.

Affume any Point F in the Right Line DF; from which let FG be drawn perpendicular to the Planet of this.

paffing