BG is perpendicular to both the Planes drawn thro' A B, BC, and DE, EF. But those Planes to which the same Right Line is perpendicular, are * parallel; I 14 of tbis. therefore the Plane drawn thro' AB,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 through those Right Lines are parallel to each other' ; 'which was to be demonstrated. PROPOSITION XVI. THE OR E M. If two parallel Planes are cut by another Planes their common Se{tions will be parallel, LET two parallel Planes AB, CD, be cut by any Plane EFGH; and let their common Sections be E F, GH. I say, E F 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 FH, and meet in K ; then, because E F K is in the Plane A B, all Points taken in E FK will be in the same Plane. But K is one of the Points that is in EFK; therefore K is in the fame Plane A B. For the fame Reason, K is also in the Plane CD; wherefore the Planes A B, CD, will meet each other. But they do not meet, fince they are supposed parallel; therefore the Right Lines E'F, GH, will not meet on the Side FH. After the same manner it is proved, that they will not meet, if produced on the Side 'E G. But Right Lines, that will neither way meet each other, are parallel ; therefore E F is parallel to GH. If, therefore, two parallel Planes are cut by any other Plane, their common Sellions will be parallel; which was to be demonstrated, '79. PROPOSITION XVII. THEOR E M. ibey fall be cut in the fame Proportion. LET two Right Lines A B, CD, be cut by parallel Planes G H, KL, MN, in the Points A, E, B, C, F, D. I say, as the Right Line A E is to the Right Line E B, so is C F to F D. For, let A C, 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 K L M N, are cut by the Plane E B D X, their common Sections • 16 of tbis. E X, BD, are * parallel.' For the same Reason, be cause two parallel Planes GH, KL, are cut by the Plane AXFC, their common Sections AC, FX, are parallel; and because E X is drawn parallel to the Side B D of the Triangle ABD, it hall be, as † 2. 6. A E is to E B, so is # A X to XD. Again, becaufe THE.ORE M. p. iben all Planes palsing through that Line will be perpendicular to the same Plane. L ET the Right Line A B be perpendicular to the Plane CL. I say, all Planes that pass through A B, are likewise perpendicular to the Plane CL. For, let a Plane D E pass thro' the Right Line A B, whofe common Section with the Plane CL, is the Right Line CE; and take some Point F in CE: from which let F G be drawn in the Plane D Е, perpendi | 11. 5. cular to the Right Line CE: Then, because A B is perpendicular to the Plane CL, it shall also be * per. Def. 3. pendicular to all the Right Lines which touch it, and are in the same Plane: Wherefore, it is perpendicular to CE; and, consequently, the Angle A BF is a Right Angle : But G F B is likewise a Right Angle ; therefore À B is parallel to F G..But' A B is at Right Angles to the Plane CL; therefore F G will be † at +8 of this. 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 Section of the Planes, are I perpendicular to the other t Def. 4. of Plane. But FG is drawn in one Plane D Е, per- ibis. pendicular 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 Е, 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 same manner we demonstrate, that all Planes paffing thro' the Right Line A B, are perpendicular to the Plane CL. Therefore, if a Right Line be perpendicular to some Plané, then all Planes paling through that Line, will be perpendicular to the fame Plane ; which was to be démonstrated. PROPOSITION XIX. THEOREM. if two Planes, cutting each otber, be perpendicu lar 10 some Plane, then their common Section will be perpendicular 10 that same Plane. LET two Planes A B, B.C, cutting each other, be perpendicular to some third Plane, and let their common Section be BD. I say, B D is perpendicular to the said third Planel which let be ADC. For, if possible, let B D not be perpendicular to the third Plane ; and from the Point D, let D E be drawn in the Plane A B, perpendicular to AD; and let D F be drawn in the Plane BC, 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 AD; DE thall |