Def. 4. fhall be perpendicular to the third Plane ADC. In like manner we prove, that D F alfo is perpendicular to the faid Plane; wherefore two Right Lines ftand at Right Angles to this third Plane, on the fame Side, at +13 of bis, the fame Point D; which is † abfurd: Therefore, to this third Plane cannot be erected any Right Lines perpendicular at D, and on the fame Side, except B D, the common Section of the Planes A B, BC: Wherefore D B is perpendicular to the third Plane. If therefore, two Planes cutting each other, be perpendicular to fome Plane, then their common Section will be perpendicular to that fame Plane; which was to be demonftrated. 23. 7. † 4. I. $25. 1. PROPOSITION XX. THEOREM. If a folid Angle be contained under three plane Angles, any two of them, howsoever taken, are greater than the third. LET the folid Angle A be contained under three plane Angles BAC, CAD, DA B. I fay, any two of the Angles BAC, CAD, DA B, are greater. than the third, howfoever taken. For, if the Angles BAC, CAD, DA B, be equal, it is evident, that any two, howfoever taken, are greater than the third; but, if not, let B A C be the greater, and make the Angle BAE at the Point A, with the Right Line A B, in a Plane paffing thro' BA, A C, equal to the Angle D A B ; make AE equal to A D; thro' E draw BEC, cutting the Right Lines AB, A C, in the Points B, C; and join D B, DC: Then, becaufe DA is equal to A E, and A B is common, the two Sides D A, AB, are equal to the two Sides A E, AB; but the Angle DA B, is equal to the Angle BAE; therefore the Bafe D B is + equal to the Bafe BE: And fince the two Sides D B, DC, are greater than BC, and DB has been proved equal to BE;' therefore the remaining Side D C fhall be greater than the remaining Side EC; and fince DA is equal to A E, and AC is common, and the Bafe DC greater than the Bafe EC; the Angle DAC fhall be greater than than the Angle EAC. But from the Construction, the Angle D A B, is equal to the Angle BAE; where fore the Angles DA B, DA C, are greater than the Angle B AC. After this manner we demonftrate, if any two other Angles be taken, that they are greater than the third. Therefore, if a folid Angle be contained under three plane Angles, any two of them, howsoever taken, are greater than the third; which was to be demonftrated. PROPOSITION XXI. THEOREM. Every Jolid Angle is contained under plane Angles, LET A be a folid Angle, contained under plane * For, take any Points, B, C, D, in each of the Lines AB, A C, AD; and join BC, CD, DB: Then, because the folid Angle at B is contained under three plane Angles CBA, ABD, CBD; any two of these are greater than the third: Therefore the Angles • 20 of this, CBA, ABD, are greater than the Angle CBD. For the fame Reason, the Angles BC A, AC D, are greater than the Angle BCD; and the Angles CD A, AD B, greater than the Angle CD B. Wherefore the fix Angles CBA, ABD, BCA, ACD, CD a, AD B, are greater than the three Angles CB D, BCD, CD B. But the three Angles CBD, BCD, CD B, are equal to two Right Angles; +32. i wherefore the fix Angles C BA, ABD, BCA, ADC, DCA, AD B, are greater than two Right Angles. And fince the three Angles of each of the Triangles ABC, A CB, A D B, are equal to two Right Angles, the nine Angles of thofe Triangles CBA, BCA, BAC, ACD, CAD, ADC, ADB, ABD, DA B, are equal to fix Right Angles; fix of which Angles CBA, BCA, ACD, A DC, AD B, ABD, are greater than two Right Angles. Therefore the three other Angles B AC, CAD, DA B, which P † 24. 3.. which contain the folid Angle, will be lefs than four PROPOSITION XXII. If there be three plane Angles, whereof two, any LET ABC, DEF, GHK, be given plane An gles, any two whereof are greater than the third; and let the equal Right Lines A B, 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 A C, DF, GK; that is, any two of them, howloever 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 A C will be greater than either D F, or GK; and it is manifeft, that A C, together with either D F, or GK is greater than the other. I fay, likewife, that 123.1.3 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 AB, BC, DE, EF, GH, HK; and join AL, CL. Then, because the two Sides A B, BL, are equal to the two Sides GH, 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 Hare greater than the Angle A B C, 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 fince the two Sides L B, BC, are equal to the two Sides, DE, E F, each to each; and the Angle DEF . DEF is greater than the Angle L BC, the Bafe D F 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 muß bé lefs than four Right Angles. L ET ABC, 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. 24. I. Let the Right Lines A B, B C, 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 this. equal to A C, DF, GK: And fo + let the Triangle † 22. 4. LMN be made, fo that A C be equal to L M, and DF to M N, and G K to LN; and let the Circle LMN be described about the Triangle, whose Cen- ‡ 5. 4° tre 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 L X, M X, NX: I fay, AB is greater than L X. For, if this be not fo, A B fhall be either equal to L X, or lefs. First, let it be equal; then because A B is equal to LX, and alfo to BC, LX fhall be equal to B C: 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 Bafe LM; wherefore the Angle A B C fhall be equal to the Angle L X M. * 8. 1. P 2 * For Cor. 15.1. +2. 6. $4.6. For the fame Reafon, the Angle DEF is equal to the. Angle M X N, and the Angle GHK to the Angle NXL; therefore the three Angles ABC, DEF, GHK, are equal to the three Angles L X M, M X N, NXL. But the three Angles L X M, M XN, NXL * are equal to four Right Angles; and fo the three Angles ABC, DE F, 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 lefs than LX; for, if this be poffible, make X O equal to B A, and XP to B C, and join OP: Then, because A B is equal to BC, XO fhall be equal to X P; and the remajning Part O L, equal to the remaining Part PM; and fo L M is † parallel to O P, and the Triangle LMX is equiangular to the Triangle OP X: Wherefore X List 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 LM is put equal to A C ; wherefore AC fhall be greater than OP: And fo, because the two Right Lines A B, B C, are equal to the two Right Lines OX, X P, and the Bafe AC greater than the Bafe OP; the Angle A B C will be greater than the Angle O X P. In like manner we demonftrate, that the Angle DEF is greater than the Angle MXN, and the Angle GHK than the Angle NXL; therefore the three Angles A BC, DEF, GHK, are greater than the three A ngles LXN, MXN, NXL: But the Angles ABC, DE F, G H K, are put lefs than four Right Angles; therefore the Angles LXM, M X N, N X L, fhail be lefs by much than four Right G. 15. 1. Angles, and also equal + to four Right Angles: which is abfurd: Wherefore A B is not lefs than L X. It has also been proved not to be equal to it; therefore 12 of this. it must neceffarily be greater. On the Point X raife X R, perpendicular to the Plane of the Circle LMN, who'e Length let be fuch, that the Square thereof be qual to the Excefs by which the Square of A B exceeds the Square of LX; and let RL, R M, RN, be joined Because RX is perpendicular to the Plane Dif. 3. of the Circle L MN, it fhall alfo be * perpendicular to LX, MX, NX: And becaufe L X is equal to #25.7. * |