Book XI. 8 7. II. b 29. I. Let AB, CD be two parallel ftraight lines, and let one of them AB be at right angles to a plane; the other CD is at right angles to the fame place. Let AB, CD meet the plane in the points B, D, and join BD: Therefore 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 becaufe AB is perpendicular to the plane, it is perpendicular to every ftraight line which meets it, and is in that a 3. def. 11. plane: Therefore each of the angles ABD, ABE, is a right angle: And because the straight line BD meets the parallel ftraight lines AB, CD, the angles ABD, CDB are together equal to two right angles: And ABD is a right angle; therefore alfo CDB is a right angle, and CD perpendicular to BD: And becaufe 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 A the angle EDB, because each of them is a right angle; therefore the bafe AD is equal to the bafe BE: Again, becaufe AB is equal to DE, and BE to AD; the two AB, BE are equal to the two ED, DA; and the bafe AE is com- B mon to the triangles ABE, EDA; wherefore the angle ABE is equal to the angle EDA: And ABE is a right angle; and therefore EDA is a right angle, and ED perpendicular to DA: C 4. I. d 8. I. d E C D But it is alfo perpendicular to BD; therefore ED is perpendi= cular to the plane which paffes through BD, DA, and shall! f 3. def. 11. make right angles with every ftraight line meeting it in that € 4. II. plane: But DC is in the plane paffing through BD, DA, becaufe 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 alfo at right angles to DB; CD then is at right angles to the two straight lines DE, DB in the point of their interfection D; and therefore is at right angles to the plane paffing through DE, DB, which is the fame plane to which AB is at right angles. Therefore, if two ftraight lines, &c. Q. E. D. PROP Book XI. w Two PRO.P. IX. THEOR. WO ftraight lines which are each of them parallel to the fame ftraight line, and not in the fame plane with it, are parallel to one another. Let AB, CD be each of them parallel to EF, and not in the fame plane with it; AB fhall be parallel to CD. A H E B G a 4. II. In EF take any point G, from which draw, in the plane paffing through EF, AB, the ftraight line GH at right angles to EF; and in the plane paffing through EF, CD, draw GK at right angles to the fame EF. And because EF is perpendicular both to GH and GK, EF is perpendicular to the plane HGK paffing' through them: And EF is parallel to AB; therefore AB is at right angles to the plane HGK. For the fame reafon, CD is likewife at right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane HGK. But if two ftraight lines be at right angles to the fame plane, they fhall be parallel to one another. Therefore AB is c 6. 11. parallel to CD. Wherefore two straight lines, &c. Q. E. D. IF CK PROP. X. THEOR. D two ftraight lines meeting one another be parallel to two others that meet one another, and are not in the fame plane with the first two; the first two and the other two fhall contain equal angles. Let the two ftraight lines AB, BC which meet one another be parallel to the two ftraight lines DE, EF that meet one another, and are not in the fame 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 ÁD, b 8. II. a 33. I. B C Book XI. AD, CF, BE, AC, DF: Because BA is equal and parallel to ED, therefore AD is both equal and parallel to BE. For the fame reason, CF is equal and parallel to BE. Therefore AD and CF are each of them e- A qual and parallel to BE. But straight lines that are parallel to the fame straight line, and not in the fame plane with it, b 9. II. are parallel to one another. Therefore c 1. Ax. 1. AD is parallel to CF; and it is equal to it, and AC, DF join them towards the fame parts; and therefore AC is equal and parallel to DF. And be. D d 8. I. a 12. 1. b II. I. C 31. I. d 4. II. e 8. 11. E caufe AB, BC are equal to DE, EF, and the bafe AC to the bafe DF; the angle ABC is equal to the angle DEF. Therefore, if two ftraight lines, &c. Q. E. D. T PROP. XI. PROB. draw a ftraight 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 ftraight line perpendicular to the plane BH. A E F H In the plane draw any ftraight line BC, and from the point A draw AD perpendicular to BC. If then AD be alfo perpendicular to the plane BH, the thing required is already done; but if it be not, from the point D draw, in the plane BH, the ftraight line DE at right angles to BC; and from the point A draw AF perpendicular to G DE; and through F draw GH parallel to BC: And becaufe BC is at right angles to ED and DA, BC is at right angles to the plane paffing through ED, DA. And GH is parallel to BC; but, if B D C two ftraight lines be parallel, one of which is at right angles to a plane, the other fhall be at right angles to the fame plane; wherefore GH is at right angles to the plane through ED, DA, f 3. def. 11. and is perpendicular to every ftraight line meeting it in that plane. But AF, which is in the plane through ED, DA, meets it: it: Therefore GH is perpendicular to AF; and confequently Book XI. AF is perpendicular to GH; and AF is perpendicular to DE; therefore AF is perpendicular to each of the ftraight lines GH, DE. But if a ftraight line ftands at right angles to each of two straight lines in the point of their interfection, it shall also be at right angles to the plane paffing through them. But the plane paffing 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. O erect a ftraight line at right angles to a given T plane, Let A be the point given in the plane; it is required to erect a ftraight line from the point A at right angles to the plane. From any point B above the plane draw from A draw b AD parallel to BC. Becaufe, therefore, AD, CB are two parallel ftraight lines, and one of them BC perpendicular to it; and BC is at right angles to the given plane, the other AD is alfo at right angles to D B it. Therefore a straight line has been erected at right angles to c 8. 11. a given plane from a point given in it. Which was to be done. F ROM the fame point in a given plane, there cannot be two ftraight lines at right angles to the plane, upon the fame fide of it: And there can be but one perpendicular to a plane from a point above the plane. For, if it be poffible, let the two itraight lines AC, AB, be at right angles to a given plane from the fame point A in the plane, and upon the fame fide of it; and let a plane pafs through BA, AC; the common fection of this with the given plane is a straight Book XI. line paffing through A: Let DAE be their common fection; Therefore the ftraight lines AB, AC, DAE are in one plane: And because CA is at right angles to the given plane, it shall a 3. II. b 6. II. make right angles with every ftraight B C line meeting it in that plane. But D A E PLA PROP. XIV. THEOR. LANES to which the fame ftraight line is perpendi cular, are parallel to one another. Let the ftraight line AB be perpendicular to each of the planes CD, EF; thefe planes are parallel to one another. G If not, they fhall meet one another when produced; let them meet; their common section fhall be a straight line GH, in which take any point K, and join AK, BK: Then, because AB is perpendicular to 3. def. 11. the plane EF, it is perpendicular to the ftraight line BK which is in that plane. Therefore ABK is a right angle. For the fame reafon, BAK is a right angle; wherefore the two angles ABK, BAK of the triangle ABK are equal to two right angles, which is impofb 17. I. fible: Therefore the planes CD, EF, though produced, do not meet one anc 8. def. 11. other; that is, they are parallel ©. Therefore planes, &c. Q. E. D. II F A B E D PROP. |