PROP. IX. THEOR. Book XI. 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 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 perpendicular 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 B a 4. 11. G E F b 8. 11. D C K HGK. But if two straight lines be at right angles to the same plane, they shall be parallel 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 Book XI. fore AD is both equal and parallel to BE. For the same reason CF is equal and paa 33. 1. rallel to BE. Therefore AD and CF are each of them equal and parallel to BE. But straight lines that are parallel to the same straight line, and not in the same b 9. 11. plane with it, are parallel to one another. Therefore AD is parallel to CF; c1.Ax.1. and it is equal to it, and AC, DF join d 8. 1. them towards the same parts; and there- a 12. 1. b 11. 1. c 31. 1. 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. In the plane draw any straight line BC, and from the point A draw a 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 draw b, 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 draw GH parallel to BC: and because BC is at right angles to ED and DA, BC is at right and 4. 11. glesd to the plane passing through ÉD, 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 e 8. 11. be at right angles to the same plane; wherefore GH is at right angles to the plane through ED, 3. def. DA, and is perpendicular f to 11. A E F G H B D C every straight line meeting it in that plane. But AF, which is in the plane through ED, DA, meets it: therefore GH is per 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 stands 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. From any point B above the plane draw a BC perpendicular to it; and from A draw b 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 D B line has been erected at right angles to a given plane from a point given in it. Which was to be done. 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. Book XI. AC; the common section of this with the given plane is a straight line passing through A: let DAE be their common section: 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 b 6. 11. be parallel b to one another, which D B A C is absurd. Therefore, from the same point, &c. Q. E. D. E 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. 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 a 3. def. EF, it is perpendicular a to the straight 11. line BK which is in that plane. Therefore 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 b 17.1. right angles, which is impossible b: therefore the planes CD, EF, though produced, do not meet one another; 8. def. that is, they are parallel c. Therefore, planes, &c. Q. E. D. 11. C G Book XI. PROP. XV. THEOR. IF two straight lines meeting one another, be pa- See N. rallel to two straight lines which meet one another, but are not in the same plane with the first two; the plane which passes through these is parallel to the plane passing through the others. Let AB, BC, two straight lines meeting one another, be parallel to DE, EF, that meet one another, but are not in the same plane with AB, BC: the planes through AB, BC, and DE, EF shall not meet, though produced. From the point B draw BG perpendiculara to the plane a 11. 11. which passes through DE, EF, and let it meet that plane in G; and through G draw GH parallel to ED, and GK pa- b 31. 1. rallel to EF and because BG is perpendicular to the plane through DE, EF, it shall make right angles with every E gles BGH, BGK is a right an gle and because BA is pa ralleld to GH (for each of A them is parallel to DE, and H d 9. 11. they are not both in the same plane with it) the angles GBA, BGH are together equal to two right angles: and BGH is a e 29. 1. right angle; therefore also GBA is a right angle, and GB perpendicular to BA for the same reason, GB is perpendicular to BC: since therefore the straight line GB stands at right angles to the two straight lines BA, BC, that cut one another in B, GB is perpendicular to the plane through BA, BC; and it is f 4. 11. perpendicular to the plane through DE, EF: therefore BG is perpendicular to each of the planes through AB, BC, and DE, EF: but planes to which the same straight line is perpendicular, are parallel g to one another: therefore the plane through AB, g 14. 11. BC is parallel to the plane through DE, EF. Wherefore, if two straight lines, &c. Q. E. D. |