Elements of geometry and trigonometry . will it also be per-pendicular to QP. Having drawn any line BC inthe plane PQ. through the lines ABand BC, draw a plane ABC. inter-secting the plane MX in AD : theintersection AD will be parallel to BC (Prop. X.) : but the lineAB, being perpendicular to the plane MX, is perpendicular tothe straight line AD ; therefore also, to its parallel BC (BookI. Prop. XX. Cor. 1.): hence the line AB being perpendicularto any line BC, drawn through lis foot in the plane PQ, is con-sequently perpendicular to that plane (Def. 1.). 31 1 ? \ T TT \ . \ \ ^ A PROPOSITION
Elements of geometry and trigonometry . will it also be per-pendicular to QP. Having drawn any line BC inthe plane PQ. through the lines ABand BC, draw a plane ABC. inter-secting the plane MX in AD : theintersection AD will be parallel to BC (Prop. X.) : but the lineAB, being perpendicular to the plane MX, is perpendicular tothe straight line AD ; therefore also, to its parallel BC (BookI. Prop. XX. Cor. 1.): hence the line AB being perpendicularto any line BC, drawn through lis foot in the plane PQ, is con-sequently perpendicular to that plane (Def. 1.). 31 1 ? \ T TT \ . \ \ ^ A PROPOSITION XII. THEOPvEM. 77ie parallels comprcheTided between Vjoû parallel planes are equal. Let MX. PQ. be two parallelplanes, and FH. GE. two paral-lel lines : then will EG=FH. For, through the parallels EG,FH, draw the plane EGHF, in-tersecting the parallel planes inEF and GH. The intersections EF, GH, are parallel to eachother (Prop. X.) : so likewise are EG. FH ; therefore the figureEGHF is a parallelogram ; con-sequently, EG = FH. 31 E. H Cor. Hence it follows, that /xo parallel planes are everywhere equidistant : for, suppose EG were perpendicular to theplane PQ ; the parallel FH would also be perpendicular to it(Prop. VH.), and the two parallels would likewise be perpen-dicular to the plane MX (Prop. XI.) : and being parallel, theywill be equal, as shown by the Proposition. BOOK VI. 135 PROPOSITION XIII. THEOREM. If two angles, not situated in the same plane, have their sidesparallel and lying in the Sfirne direction, those angles will beequal and their planes will be parallel. Let the angles be CAE and DBF. Make AC = BD, AE= M BV^ ; and draw CE, DF,AB, CD, EF. Since ACis equal and parallel toBD, the figure ABDC isa parallelogram ; thereforeCD is equal and parallelto AB. For a similar rea-son, EF is equal and par-allel to AB ; hence also CDis equal and parallel toEF ; hence the figureCEFD is a parallelogram,and the side CE is equaland paiallel to DF; therefore the triangles CAE
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