to DE, and BE to AD; AB, BE are equal to ED, DA; and, Book XI. in the triangles ABE, EDA, the bafe AE is common; there n fore the angle ABE is equal to the angle EDA: But ABE is c 8. 1. a right angle; therefore EDA is alfo a right angle, and ED perpendicular to DA: But it is alfo perpendicular to each of the two BD, DC: Wherefore ED is at right angles to each of the three ftraight lines BD, DA, DC in the point in which they meet: Therefore thefe three ftraight lines are all in the fame planed: But AB is in the plane in which are BD, DA, d 5. 11. because any three ftraight lines which meet one another are in one plane: Therefore AB, BD, DC are in one plane: And € 2. 11. each of the angles ABD, BDC is a right angle; therefore AB is parallel f to CD. Wherefore, if two ftraight lines, &c. Q. E. D. f 28. 1. IF PROP. VII. THEOR. two ftraight lines be parallel, the ftraight line drawn See N, from any point in the one to any point in the other, is in the fame plane with the parallels. A Let AB, CD be parallel ftraight lines, and take any point E in the one, and the point F in the other: The straight line which joins E and F is in the fame plane with the parallels. If not, let it be, if poffible, above the plane, as EGF; and in the plane ABCD in which the parallels are, draw the ftraight line EHF from E to F; and fince EGF alfo is a ftraight line, the two fraight lines EHF, EGF include a space between them, which is impoffible. Therefore the ftraight line joining the points E, F is not above the E B G H C F a 10, Ax. D plane in which the parallels AB, CD are, and is therefore in that plane. Wherefore, if two ftraight lines, &c. Q. E. D. IF two ftraight lines be parallel, and one of them is at see N, right angles to a plane; the other also shall be at right angles to the fame plane. Let 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 plane. 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 because AB is perpendicular to the plane, it is perpendicular to every ftraight line which meets it, and is in a 3. def. 11. that plane: Therefore each of the angles ABD, ABE, is a right angle: And because the ftraight line BD meets the parallel ftraight lines AB, CD, the angles AD, 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 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, becaufe 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 base AE is common 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: £ 4. I. d 8. L. e 4. II. e B E C D But it is alfo perpendicular to BD; therefore ED is perpendicu lar to the plane which paffes through BD, DA, and thall make £ 3. def. 11. right angles with every ftraight line meeting it in that 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 ftraight 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 straight lines, &c. Q. E. D. PROP PROP. IX. THEOR. Book XI. TW WO straight 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. AH B 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 becaufe 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 reason, 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. E CK a 4. 11. G F D b 8. Is. Therefore AB is c 6. 11. parallel to CD. Wherefore two ftraight lines, &c. Q. E. D. PROP. X. THEOR. IF 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 AD, Book XI. AD, CF, BE, AC, DF: Because BA is equal and parallel to ~ ED, therefore AD is both equal and a 33. I. b 9. II. B C parallel to BE. For the fame reason, CF is equal and parallel to BE. Therefore AD and CF are each of them equal and parallel to BE. But straight lines that are parallel to the fame straight line, and not in the fame plane with it, are parallel to one another. Therefore CI. Ax. I. 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 because AB, BC are equal to DE, EF, and the base AC to the bafe DF; the angle ABC is equal to the angle DEF. Therefore, if two straight lines, &c. Q. E. D. d 6. 1. a 12. 1. b II. I. C 31. I. T d D PROP. XI. PROB. E F O 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 also per pendicular 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 thro' F draw GH parallel to BC: And because 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 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, £ 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 €4.11. e 8. II. B D C t: Therefore GH is perpendicular to AF; and confequently Book XI. AF is perpendicular to GH; and AF is perpendicular to DE; herefore AF is perpendicular to each of the straight lines GH, DE. But if a ftraight line ftands at right angles to each of two ftraight lines in the point of their interfection, it fhall alfo 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, P R O B. O erect a ftraight line at right angles to a given Let A be the point given in the plane; it is required to erect From any point B above the plane draw BC perpendicular to it; and from A drawb AD parallel to BC. Becaufe, therefore, AD, CB are two parallel ftraight lines, and one of them BC is at right angles to the given plane, the other AD is alfo at right angles to D B A 'C it c. 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. FR ROM the fame point in a given plane, there cannot be two straight 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 ftraight 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 pass through BA, AC; the common fection of this with the given plane is a straight |