Book XI. g 7. 11. 13. def. 11. b 29. 1. c 4. 1. d 8.1. e 4. 11. f 3. def. 11. Let AB, CD be two parallel straight lines, and let one of them AB be at right angles to a plane: the other CD is at right angles to the same plane. C Let AB, CD meet the plane in the points B, D, and join BD: therefores AB, CD, BD are in one plane. In the plane to which AB is at right angles, draw DE at right angles to BD, and make DE equal to AB, and join BE, AE, AD. And because AB is perpendicular to the plane, it is perpendicular to every straight line which meets it, and is in that plane: therefore each of the angles ABD, ABE is a right angle: and because the straight line BD meets the parallel straight lines AB, CD, the angles ABD, CDB are together equal to two right angles: and ABD is a right angle; therefore also CDB is a right angle, and CD perpendicular to BD: and because AB is equal to DE, and BD common, the two AB, BD, are equal to the two ED, DB, and the angle ABD is equal to the angle EDB, because each of them is 'a A right angle; therefore the base AD is equal to the base BE: again, because AB is equal to DE, and BE to AD; the two AB, BE are equal to the two ED, DA; and the base AE is common to the triangles ABE, EDA; wherefore the angle ABE is equal to the angle EDA: and B ABE is a right angle; and therefore EDA is a right angle, and ED perpendicular to DA: but it is also perpendicular to BD; therefore ED is perpendiculare to the plane which passes through BD, DA, and shall make right angles with every straight line meeting it in that plane: but DC is in the plane passing through BD, DA, because all three are in the plane in which are the parallels AB, CD; wherefore ED is at right angles to DC; and therefore CD is at right angles to DE: but CD is also at right angles to DB; CD then is at right angles to the two straight lines DE, DB in the point of their intersection D; and therefore is at right anglese to the plane passing through DE, DB, which is the same plane to which AB is at right angles. Therefore, if two straight lines, &c. Q. E. D. E D 205 Book XI. PROP. IX. THEOR. TWO straight lines which are each of them parallel to the same straight line, and not in the same plane with it, are parallel to one another. Let AB, CD be each of them parallel to EF, and not in the same plane with it; AB shall be parallel to CD. A H B a 4. 11. G F b 8. 11. In EF take any point G, from which draw, in the plane passing through EF, AB, the straight line GH at right angles to EF; and in the plane' passing through EF, CD, draw GK at right angles to the same EF. And because EF is perpendicular both to 'GH and GK, EF is perpendiculara to the plane HGK passing through them: and EF is parallel to AB; therefore AB is at right angles to the plane HGK. For the same reason, CD is likewise at right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane C K D HGK. But if two straight lines be at right angles to the same plane, they shall be parallele to one another. Therefore AB is c 6. 11. parallel to CD. Wherefore, two straight lines, &c. Q. E. D. PROP. X. THEOR. IF two straight lines meeting one another be parallel to two others that meet one another, and are not in the same plane with the first two, the first two and the other two shall contain equal angles. Let the two straight lines AB, BC which meet one another be parallel to the two straight lines DE, EF that meet one another, and are not in the same plane with AB, BC. The angle ABC is equal to the angle DEF. Take BA, BC, ED, EF all equal to one another; and join AD, CF, BE, AC, DF: because BA is equal and parallel to ED, there a 33. 1. 9. 11. Book XI. fore AD isa both equal and parallel to BE. For the same reason, CF is equal and parallel to BE. Therefore AD and CF are each of them equal and parallel to BE. A But straight lines that are parallel to the same straight line, and not in the same plane with it, are parallel to one another. Therefore AD is parallel to CF; c 1. Ax. 1. and it is equale to it, and AC, DF join `them towards the same parts; and thereforea AC is equal and parallel to DF. And because AB, BC are equal to DE, D EF, and the base AC to the base DF; the angle ABC is equal to the angle DEF. Therefore, if two straight lines, &c. Q. E. D. d 8. 1. E F a 12. 1. b 11. 1. c 31. 1. d 4. 11. e 8. 11. f 3. def. 11. PROP. XI. PROB. TO draw a straight line perpendicular to a plane, from a given point above it. Let A be the given point above the plane BH; it is required to draw from the point A a straight line perpendicular to the plane BH. E A In the plane draw any straight line BC, and from the point A drawa AD perpendicular to BC. If then AD be also perpendicular to the plane BH, the thing required is already done; but if it be not, from the point D drawb, in the plane BH, the straight line DE at right angles to BC: and from the point A draw AF perpendicular to DE; and through F drawe GH parallel to BC: and because BC is at right angles to ED and DA, BC is at right angles to the plane passing through ED, DA. And GH is parallel to BC; but, if two straight lines be parallel, one of which is at right angles to a plane, the other shall be at right angles to the same plane; wherefore GH is at right angles to the plane through ED, DA, and is perpendicular to every straight line meeting it in that plane. But AF, which is in the plane through ED, DA, meets it: therefore GH is per G B F H D C pendicular to AF; and consequently AF is perpendicular to GH; Book XI. and AF is perpendicular to DE: therefore AF is perpendicular to each of the straight lines GH, DE. But if a straight line stand at right angles to each of two straight lines in the point of their intersection, it shall also be at right angles to the plane passing through them. But the plane passing through ED, GH is the plane BH; therefore AF is perpendicular to the plane BH; therefore, from the given point A, above the plane BH, the straight line AF is drawn perpendicular to that plane. Which was to be done. PROP. XII. PROB. TO erect a straight line at right angles to a given plane, from a point given in the plane. Let A be the point given in the plane; it is required to erect a straight line from the point A at right angles to the plane. D B a 11. 11. b 31. 1. From any point B above the plane drawa BC perpendicular to it; and from A drawb AD parallel to BC. Because, therefore, AD, CB are two parallel straight lines, and one of them BC is at right angles to the given plane, the other AD is also at right angles to it. Therefore a straight line has been erected at right angles to a given plane from a point given in it. Which was to be done. A C c 8, 11. PROP. XIII. THEOR. FROM the same point in a given plane, there cannot be two straight lines at right angles to the plane, upon the same side of it; and there can be but one perpendicular to a plane from a point above the plane. For, if it be possible, let the two straight lines AC, AB be at right angles to a given plane from the same point A in the plane, and upon the same side of it; and let a plane pass through BA, a 3. 11. B C Book XI. AC; the common section of this with the given plane is a straighta line passing through A: let DAE be their common sectión: therefore the straight lines AB, AC, DAE are in one plane: and because CA is at right angles to the given plane, it shall make right angles with every straight line meeting it in that plane. But DAE, which is in that plane, meets CA; therefore CAE is a right angle. For the same reason BAE is a right angle. Wherefore the angle CAE is equal to the angle BAE; and they are in one plane, which is impossible. Also, from a point above a plane, there can be but one perpendicular to that plane; for, if there could be two, they would be parallel to one another, b 6. 11. D A which is absurd. Therefore, from the same point, &c. Q. E. D. E a 3. def. 11. 17. 1. c 8. def. 11. PROP. XIV. THEOR. PLANES to which the same straight line is perpendicular, are parallel to one another. Let the straight line AB be perpendicular to each of the planes CD, EF; these planes are parallel to one another. G H If not, they shall meet one another when produced; let them meet; their common section shall be a straight line GH, in which take any point K, and join AK, BK: then, because AB is perpendicular to the plane EF, it is perpendicular to the straight line BK which is in that plane. There- C fore ABK is a right angle. For the same reason, BAK is a right angle; wherefore the two angles ABK, BAK of the triangle ABK are equal to two right angles, which is impossibleb: therefore the planes CD, EF, though produced, do not meet one another; that is, they are parallels. Therefore, planes, &c. Q. E. D. D E |