DE, EF. But those Planes te which the fame Right * 14 of this. Line is perpendicular, are * parallel. Therefore the Plane drawn thro' AB, BC, is parallel to the Plane drawn thro D. E, EE.. Wherefore, if two Right Lines, touching one another, be parallel to two Right Lines, touching one another, and not being in tbe fame Plane with them, the Planes draum thra thefe Right Lines are parallel to each other. PROPOSITION XVI. THEOREM. If two parallel Planes are cut by any other Plane, sbeir common Seations will be parallel. A Plane EFHG, and let their common Sections be EF, GH. I lay EF is parallel to GH. For if it is not parallel, EF, GH, being produced, will meet each other either on the Side FH, or EG. First let them be produced on the Side EH, and meet in K; then because EFK is in the Planę AB, all Points taken in EFK will be in the fame Plane. But K is one of the Points that is in EFK. Therefore K is in the Plane AB. For the fame Reafon K is also in the Plane CD. Wherefore the Planes AB, CD, will meet each other. But they do not meet, since they are supposed 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 pa: rallel; therefore EF is parallel to GH. If therefore, tgua parallel Planes are cut by any other Plane, their common Sections will be parallák;/ which was to be de monstrated. LE BD PROPOSITION XVII. THEORE M.. Mall be cut in the same Proportion. rallel Planes GH, KL, MN, in the Points A, E, B, C, F, D. I say, as the Right Line A E is to the Right Line EB, so is CF to FD. For let AC, BD, AD be joined : Let A D meet the Plane K L in the Point X, and join EX, XF, Then because two parallel Planes KL, MN, are cut by the Plane EBD X, their common Sections EX, parallel. For the same Reason, because* 16 of this two parallel Planes GH, KL, are cut by the Plane AXFC, their common Sections AC, ÉX, are parallel; and because EX is drawn parallel to the Side BD of the Triangle ABD, it shall be as AE is to EB, so is t AX to XD. Again, because XF ist 2. 6. drawn parallel to the Side AC of the Triangle ADC, it shall be t as AX is to XD, fo is CF to F D. But it has been proved, as AX is to XD, so is AE to EB. Therefore, as AE is to EB, so is I CF to II. 5. FD. Wherefore, if two Right Lines are cut by parallel Planes, they shall be cut in the same Proportion; which was to be demonstrated. are * PROPOSITION XVIII. THE O R E M. then all Planes palling thro’ that Line will be Plane CL. I say, all Planes that pass thro' AB, are likewise perpendicular to the Plane CL. For let a Plane DE pass thro' the Right Line AB, whose common Section, with the Plane CL, is the Right Line CE; and take some Point Fin CE; from which which let FG be drawn in the Plane DE, perpendi cular to the Right Line CE. Then because AB is • Def. 3. perpendicular to the Plane CL, it shall also be * per pendicular to all the Right Lines which touch it, and are in the fame Plane. Wherefore it is perpendicu lar to CE; and consequently the Angle ABF is a Right Angle; but GFB is likewise a Right Angle. Therefore AB is parallel to FG. But A B is at Right + 8 of this. Angles to the Plane CL. Therefore FG will be + at Right Angles to that same Plane. But one Plane is perpendicular to another, when the Right Lines, drawn in one of the Planes perpendicular to the common I Def4 Section of the Planes, are I perpendicular to_the other of this, Plane. But FG is drawn in one Plane DE, perpendicular to the common Section CE of the Planes. And it has been proved to be perpendicular to the Plane CL. Therefore the Plane DE is at Right Angles to the Plane CL. After the fame manner it is demonstrated, that all Planes, paffing thro' the Right Line A B, are perpendicular to the Plane CL, Therefore, if a Right Line be perpendicular to fome Plane, then all Planes pafsing thro that Line will be perpendicular to the fame Plane; which was to be demonstrated. PROPOSITION XIX. THEOREM. to fome Plane, then their common Section will be be perpendicular to some third Plane, and let their common Section be BD. I say, BD is perpendicular to the said third Plane, which let be ADC. For, if possible, let B D not be perpendicular to the third Plane ; and from the Point D, let DE be drawn in the Plane AB, perpendicular to AD; and let DF be drawn in the Plane BC, perpendicular to CD; then because the Plane AB is perpendicular to the third Plane, and DE is drawn in the Plane AB, perpendicular to their common Section AD, DE Ihall shall be * perpendicular to the third Plane. In like* Def. 4manner we prove, that DF also is perpendicular to the faid Plane. Wherefore two Right Lines stand at Right Angles, to this third Plane, on the fame Side at the fame Point D, which ist absurd. Therefore tot 13 of tbisa 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 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 same Plane ; which was to be demonstrated. PROPOSITION XX. THEOR IM. gles, any two of them, bowsoever taken, Àre. 'T LE ET the solid Angle A be contained under three plane Angles BAC, CAD, DAB, I say any two of the Angles BAC, CAD, DAB, are greater than the third, howsoever taken. For if the Angles B AC, CAD, DAB, be equal, it is evident that any two, howsoever taken, are greater than the third. But if not, let BAC be the greater ; and make the Angle B AE, at the Point A, 23. 1. with the Right Line A B, in a Plane passing thro' BA, AC, equal to the Angle DAB, make A E equal to AD;, thro' E draw BEC, cutting the Right Lines, AB, AC, in the Points B, C, and join DB, DC. Then because DA is equal to A E, and AB is common, the two Sides DA, AB, are equal to the two Sides A E, AB; but the Angle D'AB is equal to the Angle BAE. Therefore the Base DB is t equal to 1 4. 19 the Base BE. And since the two Sides DB, DC, are greater than BC, and DB has been proved equal to BE, the remaining Side DC fall be greater than the remaining Side EC; and since DA is equal to AE, and AC is common, and the Base DC greater than the Base E C, the Angle DAC fhall be I greater ( 25. 1, than Reason than the Angle E'AC. But from Construction, the THE ORE M. together less than four Right ones. ET A be a solid Angle, contained under plane Angles B AC, CAD, DAB. I say the Angles BAC, CAD, DAB, are less than four Right Ăngles. For take any Points B, C, D, in each of the Lines AB, AC, AD, and join BC, CD, D B. Then because the folid Angle at B is contained under three plane Angles CBA, ABD, CBD, any two of these * 20 of ilus. are * greater than the third. Therefore the Angles CBA, ABD, are greater than the Angle CBD. For , the Angles BCA, A CD, are greater than the Angle BCĎ; and the Angle CDA, ADB, greater than the Angle CDB. Wherefore the fix Angles CBA, ABD, BCA, ACD, ADC, ADB, are greater than the three Angles CBD, BCD, CDB. But the three Angles CBD, BCD, CDB, . 32. 1. are + equal to two Right Angles. "Wherefore the fix Angles CBA, ABD, BCA, ACD, ADC, ADB, are greater than two Right“Angles. And since the three Angles of each of the Triangles ABC, ACD, ADB, 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. Six of which Angles CBA, BCA, ACD, ADC, ADB, ABD, are greater than two Right Angles. Therefore the three other Angles BAC, CAD, DAB, which contain the folid An gle, |