* BG is perpendicular to both the Planes drawn thro' A B, BC, and D E, EF. But thofe Planes to which the fame Right Line is perpendicular, are parallel; * 14 of this therefore the Plane drawn thro' A B, BC, is parallel to the Plane drawn thro' DE, EF. Wherefore, if two Right Lines, touching one another, be parallel to two Right Lines touching one another, and not being in the fame Plane with them; the Planes drawn thro' thofe Right Lines are parallel to each other; which was to be demonftrated. PROPOSITION XVI. THEOREM. If two parallel Planes are cut by another Plane, their common Sections will be parallel. LET two parallel Planes A B, CD, be cut by any Plane EFGH; and let their common Sections be EF, GH. I fay, EF is parallel to G H. For, if it is not parallel, E F, G H, being produced, will meet each other either on the Side FH, or EG. First, let them be produced on the Side FH, and meet in K; then, because E F K is in the Plane A B, all Points taken in E F K will be in the fame Plane. But K is one of the Points that is in EF K; therefore K is in the fame Plane A B. For the fame Reason K is alfo in the Plane CD; wherefore the Planes A B, CD, will meet each other. But they do not meet, fince they are fuppofed parallel; therefore the Right Lines EF, GH, will not meet on the Side FH. After the fame manner it is proved, that they will not meet, if produced, on the Side EG. But Right Lines, that will neither Way meet each other, are parallel; therefore EF, is parallel to GH. If, therefore, two parallel Planes are cut by any other Plane, their common Sections will be parallel; which was to be demonftrated. PRO PROPOSITION XVII. THEOREM. If two Right Lines are cut by parallel Planes, they shall be cut in the fame Proportion. LE ET two Right Lines A B, CD, be cut by parallel Planes GH, KL, MN, in the Points A, E, B, C, F, D. I fay, as the Right Line A É is to the Right Line E B, fo is C F to FD. For, let AC, BD, AD, be joined; let AD meet the Plane KL in the Point X; and join E X, XF. Then, because two parallel Planes KL, MN, are cut by the Plane EBDX, their common Sections * 16 of this. E X, BD, are * parallel. For the fame Reason, because two parallel Planes GH, KL, are cut by the Plane A XFC, their common Sections A C, FX, are parallel; and becaufe EX is drawn parallel to the Side BD of the Triangle A B D, it fhall be, as AE is to EB, fo is + AX to XD. Again, because XF is drawn parallel to the Side AC of the Triangle ADC, it fhall be †, as A X is to XD, fo is C F to FD. But it has been proved, as A X is to X D, fo is AE to EB. Therefore, as AE is to EB, fo is t CF to FD. Wherefore, if two Right Lines are cut by parallel Planes, they shall be cut in the fame Proportion; which was to be demonstrated. + 2.6. 11. 5. PROPOSITION XVIII. THEOREM. If a Right Line be perpendicular to fome Plane, then all Planes passing thro' that Line will be perpendicular to the fame Plane. ET the Right Line AB be perpendicular to the Plane CL. I fay, all Planes that pafs thro' A B, are likewife perpendicular to the Plane CL. For, let a Plane D E pafs thro' the Right Line A B, whofe common Section, with the Plane CL, is the Right Line CE; and take fome Point Fin CE; from which let F G be drawn in the Plane D E, perpendi cular this. cular to the Right Line CE: Then, becaufe A Bis perpendicular to the Plane CL, it fall alfo be per- * Def. 3. pendicular to all the Right Lines which touch it, and are in the fame Plane: Wherefore it is perpendicular to CE; and, conféquently, the Angle ABF is a Right Angle: But GFB is likewife a Right Angle; therefore A B is parallel to F G. But A B is at Right Angles to the Plane CL; therefore F G will be tatt 8 of thise Right Angles to that fame Plane. But one Plane is perpendicular to another, when the Right Lines drawn in one of the Planes, perpendicular to the common Section of the Planes, are ‡ perpendicular to the other † Def 4 of Plane. But FG is drawn in one Plane D E, perpendicular to the common Section CE of the Planes, and it has been proved to be perpendicular to the Plane CL: In like manner any other Line in the Plane D E, drawn perpendicular to CE, is proved to be perpendicular to the Plane CL. Therefore the Plane DE is at Right Angles to the Plane CL. After the fame manner we demonftrate, that all Planes. paffing thro' the Right Line A B, are perpendicular to the Plane C L. Therefore, if a Right Line be perpendicular to fome Plane, then all Planes, paffing thro' that Line, will be perpendicular to the fame Plane; which was to be demonftrated. PROPOSITION XIX. THEOREM. If two Planes, cutting each other, be perpendicu lar to fome Plane, then their common Section will be perpendicular to that fame Plane. L ET two Planes AB, BC, cutting each other, be 'perpendicular to fome third Plane, and let their common Section be BD. Ifay, BD is perpendicular to the faid third Plane, which let be A D C. For, if poffible, let BD not be perpendicular to the third Plane; and from the Point D let DE be drawn in the Plane A B, perpendicular to AD; and let DF be drawn, in the Plane B C, perpendicular to CD: Then, because the Plane A B is perpendicular to the third Plane, and D E is drawn in the Plane A B, perpendicular to their common Section A D; DÉ fhall 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 t13 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 BD, the common Section of the Planes AB, BC: Wherefore DB is perpendicular to the third Plane. If there. fore, 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. 23.1. †4. I. $ 25. I. PROPOSITION XX. THEOREM. If a folid Angle be contained under three plane Angles, any two of them, bowfoever taken, are greater than the third. LE ET the folid Angle A be contained under three plane Angles B AC, CAD, DAB. I fay, any two of the Angles BAC, CAD, DA B, are greater than the third, howsoever taken. For, if the Angles BAC, CAD, DAB, be equal, it is evident, that any two,howfoever taken, are greater than the third; but, if not, let BAC be the greater, and make the Angle B A E, at the Point A, with the Right Line A B, in a Plane paffing thro' B A, AC, equal to the Angle DAB; make A E equal to AD; thro' E draw BEC, cutting the Right Lines A B, A C, in the Points B, C; and join D B, DC: Then, because DA is equal to AE, and AB is common, the two Sides DA, A B, are equal to the two Sides AE, AB; but the Angle DAB, is equal to the Angle BAE; therefore the Bafe D B is † equal to the Base BE: And fince the two Sides D B, DC, are greater than BC, and D B has been proved equal to BE; therefore the remaining Side DC 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 Conftruction, the Angle DA B, is equal to the Angle BAE; wherefore the Angles DAB, DAC, are greater than the Angle BAC. 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. THE ORE M. Every folid Angle is contained under plane Angles, together, less than four Right ones. LE ET A be a folid Angle, contained under plane Angles BAC, CAD, DAB. I fay, the Angles BAC, CAD, DA B, are less than four Right Angles. For, take any Points, B, C, D, in each of the Lines A B, AC, AD; and join BC, CD, DB: Then, because the folid Angle at B is contained under three plane Angles C BA, 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 B C A, ACD, are greater than the Angle BCD; and the Angles CDA, AD B, greater than the Angle CDB. Wherefore the fix Angles CBA, ABD, BCA, ACD, CDA, ADB, are greater than the three Angles CBD, BCD, CDB. But the three Angles C B D, BCD, CD B, are + equal to two Right Angles ; † 32. *. wherefore the fix Angles CBA, A BD, BCA, ADC, DCA, AD B, are greater than two Right Angles. And fince the three Angles of each of the Triangles ABC, ACB, AD B, are equal to two Right Angles, the nine Angles of thofe Triangles CBA, BCA, BAC, ACD, CAD, ADC, ADB, ABD, DAB, are equal to fix Right Angles; fix of which Angles CBA, BCA, ACD, ADC, ADB, ABD, are greater than two Right Angles. Therefore the three other Angles B AC, CAD, DA B, P which |