gle, will be less than four Right Angles. Wherefore, every solid Angle is fontained under Angles together, less than four plane right ones; which was to be demonstrated. PROPOSITION XXII. THEORE M. bow taken, are greater than the third, and tbe: gles, 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, DF, GK, be joined. I say, it is possible to make a Triangle of AC, DF, G K, that is, any two of them, howfoever taken, are greater than the third. For if the Angles at B, E, H, are equal, then AC, DF, GK, will be * equal, and any two of them. 4. 1. 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,+244 % + and it is manifest, that AC, together with either DF, or GK, is greater than the other. I say likewise, that DF, GK, together, are greater than AC. For make f at 23.5 the Point B, with the Right Line AB, the Angle ABL; equal to the Angle GHK ; and make BL equal to either AB, BC, DE, EF, GH, HK, and join AL, CL. Then, because the two Sides AB, BL, are equal to the two Sides GH, HK, each to each and they contain equal Angles, the Base AL Ihall be equal to the Base GK. And since the Angles E and H are greater than the Angle ABC, whereof the Anz gle GHK is equal to the Angle AB L, the other Angle at E shall be greater than the Angle LBC And since the two Sides L B, BC, are equal to the two Sides DE, EF, each to each, and the Angle 24. I. DEF is greater than the Angle LBC, the Bafe DF Thall be * greater than the Base LC. But GK has been proved equal to AL. Therefore DF, GK are greater than AL, LC; but AL, LC, are greater than AC. Wherefore DF, GK, shall be much greater than -AC. Therefore any two of the Right Lines AC, DF, GK, howsoever taken, are greater than the other: And so a Triangle may be made of AC, DF, GK; which was to be demonstrated. PROPOSITION XXIII. L PROBLEM. of any two, bowsoever taken, are greater than gles given, whereof any two, howsoever taken, are greater than the other, and let the said three Angles be less than four Right Angles. It is required to make a solid Angle of three plane Angles equal to ABC, DEF, ĞHK. Let the Right Lines AB, BC, DE, EF, GH, HK, be cut off equal, and join AC, DF, GK; then it is • 22 of this. poffible to make * a Triangle of three Right Lines $ 22. 1. equal to AC, DF, GK: And so let the Triangle LMN be made, so that AC be equal to LM, and DF to MN, and GK to LN; and let the Circle 15.4 LMN be described † about the Triangle, whose Center let be X, which will be either within the Triangle LMN, or on one Side thereof, or without the fame. First, let it be within, and join LX, MX, NX. I lay AB is greater than LX. For if this be not fo, AB shall be either equal to LX, or less. First, let it be equal; then because AB is equal to LX, and also to BC, LX shall be equal to BC; but LX is equal to XM. Therefore the two Sides AB, BC, are equal to the two Sides LX, XM, each to each; but the Base A C is put equal to the Bafe LM. Where fore the Angle ABC thall be * equal to the Angle LXM LXM. For the same Reason, the Angle DEF is equal to the Angle MXN, and the Angle GHK to the Angle NXL. Therefore the three Angles ABC, DEF, GHK, are equal to the three Angles LXM, MXN, N-XL. But the three Angles L XM, MXN, NXL, are equal to four Right Angles : * Cor. 85. to And so the three Angles ABC, DEF, GHK, Ihall also be equal to four. Right Angles; but they are put less than four Right Angles, which is absurd. Therefore AB is not equal to LX. I say also it is neither less than LX;' for if this be possible, make XO equal to AB, and XP to BC, and join OP. Then because AB is equal to BC, XO shall be equal to XP; and the remaining Part OL equal to the remaining Part PM: And fo LM is t parallel to OP, and the † 2.6 Triangle LMX is equiangular to the Triangle OPX. Wherefore XL is I to LM, as XO is to OP; and I 4.6, (by Alternation) as XL is to X O, so is L M to OP. But LX is greater than XO. Therefore LM fhall also be greater than OP. But LM is put equal to AC. Wherefore AC shall be greater than OP. And so because the two Right Lines AB, BC, are equal to the two Right Lines OX, XP, and the Base AC greater than the Base OP; the Angle ABC will be * greater than the Angle OXP. In like * 25. I. manner, we demonstrate that the Angle DEF is greater than the Angle MXN, and the Angle GHK, than the Angle NXL. Therefore the three Angles ABC, DEF, GHK, are greater than the three Angles LXM, MXN, NXL. But the Angles ABC, DEF, GHK, are put less than four Right Angles. Therefore the Angles L XM, MXN, NXL, Ihall be less by much than four Right Angles, and also equal † to four Right Angles ; which is absurd. + Cor. 15. 1. Wherefore AB is not less than LX. It has also been prov'd not to be equal to it. Therefore it must necessarily be greater. On the Point X raise I XR, $ 12 of tbiso perpendicular to the Plane of the Circle LMN; whofe Length let be fuch, that the Square thereof be equal to the Excess, by which the Square of AB ex- / ceeds the Square of LX; and let RL, RM, RN, be joined. Because R X is perpendicular to the Plane of the Circle LMN, it shall also be * perpendicular * Def. 34 co LX, MX, NX.. And because LX is equal to ET AF gles giv are greater t gles be less to make a foli ABC, DI Let the be cut off • 22 of this. poffible to † 22. 1. equal to A DF to 1 15.4 LMN Center F equal, or less. First, let it be equal ; N is greater than DEF; but because the the Base DF is equal to the Base MN, the XN shall be equal to the Angle DEF; but en proved greater, which is abfurd. ThereB is not equal to LX. Moreover we will hat it is not less; wherefore it shall be neceseater. And if, again, XR be raised at Right 0 the Plane of the Circle, and made equal to of that Square, by which the Square of AB the Square of LX, the Problem will be ded. Now, I say, AB is not less than LX; is possible that it can be less, make XO equal - and XP equal to B C, and join OP. Then, · AB is equal to BC, XO shall be equal to and the remaining Part OL equal to the ing Part PM; therefore LM is * parallel to * 2. 6 6 and the Triangle LMX equiangular to the rle PXO. Wherefore as t XL is to LM, Lot 4.6. to OP: And (by Alternation) as L X is to ris LM to OP; but LX is greater than XO; re L M is greater than OP; but LM is equal C; wherefore AC shall be greater than OP. in because the two Sides AB, BC, are equal to **O Sides OX, XP, each to each; and the Base greater than the Base OP; the Angle ABC be greater than the Angle O XP. So likewise I 25.15 R be taken equal to X or XP, and OR be *), we prove that the Angle GHK is greater the Angle O XR. At the Point X, with the Line LX, make the Angle LXS equal to the - ABC, and the Angle LXT equal to the An HK, and XS, XÏ, each equal to XO, and the is |