which contain the folid Angle, will be lefs than four Right Angles. Wherefore, every folid Angle is contained under plane Angles, together, left than four Right ones; which was to be demonstrated.

PROPOSITION XXI.

THEOREM.

If there be three plane Angles, whereof two, any bow taken, are greater than the third, and the Right Lines that contain them be equal; then it is poffible to make a Triangle of the Right Lines joining the equal Right Lines which form the Angles.

LE

ET ABC, DEF, GHK, be given-plane Angles, any two whereof are greater than the third; and let the equal Right Lines AB, BC, DE, EF, GH, HK, contain them; and let A C, D F, GK, be joined. I fay, it is poffible to make a Triangle of AČ, DF, GK; that is, any two of them, howsover taken, are greater than the third.

For, if the Angles at B, E, H, are equal; then A C, DF, GK, will be equal, and any two of them greater than the third; but, if not, let the Angles at B, E, H, be unequal; and let the Angle B be greater than either of the others at E, or H: Then the Right Line AC will be + greater than either DF, or GK; and it is manifeft, that AC, together with either D F, or GK, is greater than the other. I fay, likewife, that DF, GK, together, are greater than AC. For make‡, at the Point B, with the Right Line A B, the Angle ABL equal to the Angle GHK; and make B L equal to either A B, BC, DE, EF, GH, HK, and join AL, CL. Then, because the two Sides A B, BL, are equal to the two Sides G H, HK, each to each; and they contain equal Angles; the Bafe A L fhall be equal to the Bafe GK, And fince the Angle E and H are greater than the Angle ABC, the Angle GHK is equal to the Angle A BL, and therefore the other Angle at E fhall be greater than the Angle LBC. And fence the two Sides LB, BC, are equal to the two Sides DE, EE, each to each, and the Angle DEF

DEF is greater than the Angle LBC, the Bafe DF

*

fhall be greater than the Bafe LC. But GK has* 24. 1. been proved equal to A L; therefore DF, G K, are greater than AL, LC: But A L, LC, are greater than A C; wherefore D F, GK, fhall be much greater than A C. Therefore, any two of the Right Lines AC, DF, GK, howfoever taken, are greater than the third: And fo, a Triangle may be made of AC, DF, GK; which was to be demonstrated.

PROPOSITION XXIII.

PROBLEM.

To make a folid Angle of three plane Angles, whereof any two, bowfoever taken, are greater than the third; but these three Angles must be lefs than four Right Angles.

LETABC, DEF, GHK, be three plane Angles given, whereof any two, howsoever taken, are greater than the third; and let the faid three Angles be less than four Right Angles; it is required to make a folid Angle of three plane Angles equal to A B C, DEF, GHK.

Let the Right Lines AB, BC, DE, EF, GH, HK, be cut off equal; and join AC, DF, GK; then it is poffible to make a Triangle of three Right Lines 22 of thin equal to A C, D F, GK: And fo + let the Triangle+ 22. 1. LMN be made, fo that A C be equal to L M, and DF to MN, and G K to LN; and let the Circle LMN be defcribed about the Triangle, whofe Cen-† 5.4 tre let be X, which will be either within the Triangle LMN, or on one Side thereof, or without the fame.

Firft, let it be within, and join LX, MX, NX: I fay, A B is greater than LX. For, if this be not fo, AB fhall be either equal to L X, or lefs. First, let it be equal; then, because A B is equal to LX, and also to BC, LX fhall be equal to BC: But L X is equal to X M; therefore the two Sides A B, BC, are equal to the two Sides L X, XM, each to each; but the Bafe AC is put equal to the Base LM; wherefore the Angle ABC fhall be equal to the Angle LX M.. 8.4.

*

P 2

For

❤ Cor. x5.

+2.6.

4.6.

1.

*

For the fame Reafon, the Angle DEF is equal to the Angle MXN, and the Angle GHK to. the Angle N XL; therefore the three Angles A B C, DEF, GHK, are equal to the three Angles L XM, MXN, NXL. But the three Angles LX M, MXN, NXL are equal to four Right Angles; and fo the three Angles A B C, DEF, GHK, fhall be equal to four Right Angles: But they are put lefs than four Right Angles, which is abfurd; therefore A B is not equal to LX. I fay alfo, it is neither less than LX; for, if this be poffible, make XO equal to B A, and XP to BC, and join OP: Then, because A B is equal to BC, XO fhall be equal to X P; and the remaining Part O L, equal to the remaining, Part PM; and fo LM is + parallel to O P, and the Triangle LM X is equiangular to the Triangle OPX: Wherefore XL is to LM, as XO is to OP; and (by Alternation) as XL is to XO, fo is LM to OP. But LX is greater than XO; therefore LM fhall also be greater than OP. But L M is put equal to AC; wherefore A C fhall be greater than OP: And fo, because, the two Right Lines A B, BC, are equal to the two Right Lines OX, XP, and the Bafe AC greater than the Bafe OP; the Angle ABC will be greater than the Angle O XP. In like manner we demonftrate, that the Angle DEF is greater than the Angle MXN, and the Angle GHK than the Angle N XL; therefore the three Angles ABC, DEF, GHK, are greater than the three Angles LX M, MXN, NXL: But the Angles ABC, DEF, GHK, are put lefs than four Right Angles; therefore the Angles LX M, MXN, NX L, fhall be lefs by much than four Right Cor. 15. r. Angles, and alfo equal + to four Right Angles; which is abfurd: Wherefore A B is not lefs than LX. It has also been proved not to be equal to it; therefore it muft neceffarily be greater. On the Point X raife 12 of this. XR, perpendicular to the Plane of the Circle LMN, whofe Length let be fuch, that the Square thereof be equal to the Excefs by which the Square of AB exceeds the Square of LX; and let RL, RM, RN, be joined: Because R X is perpendicular to the Plane Dif. 3 of the Circle L M N, it fhall alfo be perpendicular to LX, MX, NX: And because L X is equal to

25. 1.

*

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, RN is equal R L, or RM; therefore three Right Lines R L, RM, 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 LX, the Square of A B will be equal to the Squares of LX, XR, together: But the Square of RL is equal to the Squares of LX, XR, for LXR is at 47. 1 Right Angle; therefore the Square of AB will be equal to the Square of RL; and fo A B is equal to R L. But BC, DE, EF, GH, HK, are every one of them equal to A B; and RN, or R M, equal to RL; wherefore A B, BC, DE, EF, GH, HK, are each equal to R L, RM, or RN: And fince the two Sides RL, RM, are equal to the two Sides A B. BC; and the Bafe L M is put equal to the Bafe AC; the Angle L R M fhall be ‡ equal to the Angle ABC. Fort 8. 1. the fame Reason the Angle M R N is equal to the Angle DE F, and the Angle L R 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 A B C, DEF, GHK.

Now, let the Centre of the Circle X be in one Side of the Triangle, viz. in the Side MN; and join X L. I fay, again, that A B is greater than L X. For, if it be not fo, AB will be either equal, or less than L X. First, let it be equal; then the two Sides A B, BC, are equal to the two Sides MX, LX, that is, they are equal to MN: But M N is put equal to D F ; therefore DE, EF, are equal to D F, which is * impoffible; * 20. 1. therefore A B is not equal to LX. 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 RX 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 LMN, and join LX, MX, NX: I fay, A B is greater than LX. For, if it be not, it uft either be equal, or lefs. First, let it be equal;

P 3

then

then the two Sides A B, BC, are equal to the two Sides MX, XL, each to each; and the Bafe AC is equal to the Bafe ML; therefore the Angle ABC is + equal to the Angle MXL. For the fame Reason, the Angle G H K is equal to the Angle LXN; and fo the whole Angle MXN is equal to the two Angles ABC, GHK: But the Angles ABC, GHK, are greater than the Angle DEF; therefore the Angle MXN is greater than DEF: But because the two Sides DE, EF, are equal to the two Sides M X, X N, and the Bafe DF is equal to the Bafe MN; the Angle MNX fhall be + equal to the Angle DEF: But it has been proved greater, which is abfurd; therefore A B is not equal to LX. Moreover, we will prove, that it is not lefs; wherefore it shall be neceffarily greater. And if, again, XR be raifed 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, AB is not lefs than LX: For, if it is poffible that it can be lefs, make XO equal to AB, and X P equal to BC, and join OP; then, becaufe A B is equal to B C, XO fhall be equal to XP, and the remaining Part OL equal to the remaining Part PM; 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 L X is to XO, fo is L M to OP: But LX is greater than XO; therefore L M is greater than OP; but LM is equal to A C; wherefore AC fhall be greater than OP: "And fo, because the two Sides A B, BC, are equal to the two Sides OX, XP, each to each; and the Bafe AC is greater than the Bafe OP; $25. 1. the Angle ABC fhall be greater than the Angle OXP. So, likewife, if XR be taken equal to XO, 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 XS équal to the Angle ABC, and the Angle LXT equal to the Angle GHK,and X S, XT, each equal to XO, and join OS, OT, ST; then, because the two Sides AB, BC, are equal to the two Sides O X, X S, and the Angle A B C is equal to the Angle O XS, the Bafe

2.6.

† 4. 6.

AC