angle ABD being a right angle, the angle CDB will also be a right angle (I. 25.) But fince ED is at right angles to DB, DA, it is also at right angles to the plane which paffes through them (VII. 3.); and confequently to DC (VII. Def. 2.) The line pc is, therefore, perpendicular to each of the lines DE, DB; whence it is alfo perpendicular to the plane FG (VII. 3.), as was to be fhewn. PROP. VI. THEOREM. If two right lines be parallel to the fame right line, though not in the fame plane with it, they will be parallel to each other. Let the right lines AB, CD be each of them parallel to the right line EF, though not in the fame plane with it; then will AB be parallel to CD. For take any point G in the line EF, and draw the right lines GH, GK, each perpendicular to EF (I. 11.), in the planes AF, ED of the propofed parallels : Then fince the right line EF is perpendicular to the two right lines GH, GK, at their point of intersection G, it will also be perpendicular to the plane HGK which paffes through thofe lines (VIII. 3.) And because the lines AB, EF are parallel to each other (by Hyp.), and one of them, EF, is perpendicular to the plane HCK, the other, AB, will also be perpendicular to that plane (VII, 5.) And, And, in like manner, it may be proved that the line cD is alfo 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. 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 thofe lines will be equal. B 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 rallel to BE. pa But lines which are equal and parallel to the same 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 shewn. PROP, 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 B, A, G, make AH perpendicular to AG PROP. II. THEOREM. Any three right lines which mutually interfect each other, are all in the fame plane. E Let AB, BC, CA be three right lines, which interfect each other in the points A, B, C ; then will those lines be in the fame plane. For let any plane AD pass through the points A, B, and be turned round that line, as an axis, till it pafs through the point c. Then, because the points A, C are in the plane AD, the whole line AC muft alfo be in it; or otherwise its parts would not lie in the fame direction. And, because the points B, C are also in this plane, the whole line BC muft likewife be in it; for the fame reafon. But the line AB is in the plane AD, by hypothefis; whence the three lines AB, BC, CA are all in the fame plane, as was to be shewn. COR. Any two right lines which intersect each other, are both in the fame plane; and through any three points a plane may be extended, PROP. III. THEOREM. If a right line be perpendicular to two other right lines, at their point of interfection, it will also be perpendicular to the plane which paffes through thofe lines. Let the right line AB be perpendicular to each of the two right lines BC, BD, at their point of interfection в; then will it also be perpendicular to the plane which paffes through thofe lines. For make BD equal to BC; and, in the plane which paffes through those lines, draw any right line BE; and join the points CD, AD, AE and Ac: Then because the fide BC is equal to the fide BD (by Conft.), and the perpendicular AB is common to each of the triangles ABC, ABD, the fide AD will also be equal to the fide AC (1. 4.) And fince the triangles CAD, CBD are ifofceles, the rectangle of CE, ED, together with the square of EB, is equal to the fquare of DB; and the rectangle of CE, ED together with the fquare of EA, is equal to the square of AD (II. 20.) From each of these equals, take away the rectangle of CE, ED which is common, and the difference of the |