And, in like manner, it may be proved that the line CD is also perpendicular to the plane HGK. But when two right lines are perpendicular to the fame plane, they are parallel to each other (VII. 4.); whence the line AB is parallel to the line CD, as was to be fhewn. PRO P. VII. THEOREM. If two right lines that meet each other, be parallel to two other right lines that meet each other, though not in the fame plane with them, the angles contained by those lines will be equal. D B A Let the two right lines AB, BC, which meet each other in the point B, be parallel to the two right lines De, ef which meet each other in the point E; then will the angle ABC be equal to the angle DEF. For make BA, BC, ED, EF all equal to each other (I. 3.), and join AD, CF, BE, AC and DF. Then, because BA is equal and parallel to ED (by Hyp.), AD will be equal and parallel to BE (I. 29.) And, for the fame reason, CF will also be equal and parallel to BE. But lines which are equal and parallel to the fame line, though not in the fame plane with it, are equal and parallel to each other (I. 26. and VII. 6.); whence AD is equal and parallel to CF. And fince lines which join the corresponding extremes of two equal and parallel lines are also equal and parallel (I. 29.), AC will be equal and parallel to DF. The three fides of the triangle ABC are, therefore, equal to the three fides of the triangle DEF, each to each; whence the angle ABC is equal to the angle DEF (I. 7.), as was to be fhewn. PRO P. VIII. PROBLEM. To draw a right line perpendicular to a given plane, from a given point in the plane. Let A be the given point, and BC the given plane; it is required to draw a right line from the point A that shall be perpendicular to the plane BC. Take any point E above the plane BC, and join EA; and through a draw AF, in the plane BC, at right angles with EA (I. 11.); then if EA be alfo at right angles with any other line which meets it in that plane; the thing required is done. But if not, in the plane BC, draw AG at right angles to AF (I. 11.); and in the plane EG, which paffes through the points E, A, G, make AH perpendicular to AG, AG (I. 11.), and it will also be perpendicular to the plane BC, as was required. For, fince the right line FA is perpendicular to each of the right lines AE, AG (by Conft.), it will also be perpendicular to the plane EG which paffes through those lines (VII. 3.) And because a right line which is perpendicular to a plane is perpendicular to every right line which meets it in that plane (Def. 2.), FA will be perpendicular to AH. But AG is also perpendicular to AH (by Conft.); whence AH, being perpendicular to each of the right lines FA, AG, it will also be perpendicular to the plane BC (VII. 3), as was to be fhewn. To draw a right line perpendicular to a given plane, from a given point above it. H B Let A be the given point, and BG the given plane; it is required to draw a right line from the point A that fhall be perpendicular to the plane BG. Take any right line BC, in the plane BG, and draw AD perpendicular to BC (I. 11.); then if it be alfo perpendicular to the plane BG, the thing required is done. But if not, draw DE, in the plane BG, at right angles to BC (I. 11), and make AF perpendicular to DE (I. 12.); then will AF be perpendicular to the plane BG, as was required. For, through the point F, draw the line HG parallel to the line BC (I. 27.) Then fince the right line BC is perpendicular to each of the right lines DA, DE, it will alfo be perpendicular to the plane which paffes through those lines (VII. 3.) And because the lines BC, HG are parallel to each other, and one of them, BC, is perpendicular to the plane ADF, the other, HG, will alfo be perpendicular to that plane (VII. 5.) But if a line be perpendicular to a plane it will be perpendicular to all the lines which meet it in that plane (VII. Def. 2.); whence the line HG is perpendicular to AF. And fince the line AF is perpendicular to each of the lines HG, ED, at their point of interfection F, it will also be perpendicular to the plane BG (VII. 3), as was to be shewn. PROP. X. THEOREM. Planes to which the fame right line is perpendicular, are parallel to each other. H A Let the right line AB be perpendicular to each of the planes CD, EF; then will thofe planes be parallel to each other. For if they be not, let them be produced till they meet each other; and in the line GH, which is their common fection, take any point K; and join KA, KB: Then, because the line AB is perpendicular to the plane EF (by Hyp.), it will alfo be perpendicular to the line BK, which lies in that plane (VII. Def. 2.); and the angle ABK will be a right angle. And, for the fame reafon, the line AB, which is perpendicular to the plane DC (by Hyp.), will be perpendicular to the line AK; and the angle BAK will also be a right angle. The angles ABK, BAK are, therefore, equal to two right angles, which is abfurd (I. 28.); and confequently the planes can never meet, but must be parallel to each other (VII. Def. 6.), as was to be shewn. PROP. XI. THEOREM. If two right lines which meet each other, be parallel to two other right lines which meet each other, though not in the fame plane with them, the planes which pass through thofe lines will be parallel. E DH Let the right lines AB, BC, which meet each other in B, be parallel to the right lines DE, EF, which meet each other in E, though not in the fame plane with them; then will the plane ABC be parallel to the plane DEF. |