PROPOSITION 12. PROBLEM. To draw a straight line perpendicular to a given plane from a given point in the plane. Let A be the given point in the plane XY. It is required to draw from A a st. line perp. to the plane XY. From any point B outside the plane XY draw BC perp. to the plane. XI. 11. Then if BC passes through A, what was required is done. But if not, from A draw AD par1 to BC. Then AD shall be the perpendicular required. I. 31. 1. Equal straight lines drawn to meet a plane from a point without it are equally inclined to the plane. 2. Find the locus of the foot of the perpendicular drawn from a given point upon any plane which passes through a given straight line. 3. From a given point A a perpendicular AF is drawn to a plane XY; and from F, FD is drawn perpendicular to BC, any line in that plane shew that AD is also perpendicular to BC. PROPOSITION 13. THEOREM. Only one perpendicular can be drawn to a given plane from a given point either in the plane or outside it. W X CASE I. Let the given point A be in the given plane XY; and, if possible, let two perps. AB, AC be drawn from A to the plane XY. and Let DF be the plane which contains AB and AC; let the st. line DE be the common section of the planes DF and XY. XI. 3 Then the st. lines AB, AC, AE are in one plane. And because BA is perp. to the plane XY, .. BA is also perp. to AE, which meets it in this plane; that is, the BAE is a rt. angle. Similarly, the CAE is a rt. angle. Hyp. XI. Def. 1. the 2 BAE, CAE, which are in the same plane, are equal to one another; which is impossible. two perpendiculars cannot be drawn to the plane XY from the point A in that plane. CASE II. Let the given point A be outside the plane XY. Then two perps cannot be drawn from A to the plane; for if there could be two, they would be par', XI. 6. which is absurd. Q.E.D. PROPOSITION 14. THEOREM. Planes to which the same straight line is perpendicular are parallel to one another. B Let the st. line AB be perp. to each of the planes CD, EF. Then shall the planes CD, EF be par1. For if not, they will meet when produced. If possible, let the two planes meet, and let the st. line GH be their common section. In GH take any point K; and join AK, BK. Then because AB is perp. to the plane EF, XI. 3. .. AB is also perp. to BK, which meets it in this plane; .. the planes CD, EF, though produced, do not meet : that is, they are par'. Q.E.D. PROPOSITION 15. THEOREM. If two intersecting straight lines are parallel respectively to two other intersecting straight lines which are not in the same plane with them, then the plane containing the first pair shall be parallel to the plane containing the second pair. Let the st. lines AB, BC be respectively par' to the st. lines DE, EF, which are not in the same plane as AB, BC. Then shall the plane containing AB, BC be part to the plane containing DE, EF. From B draw BG perp. to the plane of DE, EF; XI. 11. and let it meet that plane at G. Through G draw GH, GK par1 respectively to DE, EF. 1. 31. Then because BG is perp. to the plane of DE, EF, BG is also perp. to GH and GK, which meet it in that plane : XI. Def. 1. that is, each of the BGH, BGK is a rt. angle. Now by hypothesis BA is par1 to ED, and by construction GH is par1 to ED; And since the XI. 9. Similarly the BGH is a rt. angle ; Proved. CBG is a rt. angle. the Then since BG is perp. to each of the st. lines BA, BC, .. these planes are par1. PROPOSITION 16. THEOREM. XI. 14. Q.E.D. If two parallel planes are cut by a third plane, their common sections with it shall be parallel. E B Let the par planes AB, CD be cut by the plane EFHG, and let the st. lines EF, GH be their common sections with it. Then shall EF, GH be part. For if not, EF and GH will meet if produced. Then since the whole st. line EFK is in the plane AB, XI. 1. and K is a point in that line, .. the point K is in the plane AB. Similarly the point K is in the plane CD. Hence the planes AB, CD when produced meet at K ; .. the st. lines EF and GH do not meet; Hyp. I. Def. 35. Q.E.D. |