Let them meet the plane in the points B, D. Draw DE at right angles to DB, in the plane BDE, and let E be any point in it: Join AE, AD, EB. Because ABE is a right angle, AB + BE = (47. 1.) AE%, and because BDE is a right angle, BE BD2 + DE2 ; therefore ABa + BD2 + DE2 = AE2; now, AB? +BD2=AD?, because ABD is a right angle, therefore AD? +DE = AE, and ADE is therefore a (48. 1.) right angle. There B В D fore ED is perpendicular to the three lines BD, DA, DC, whence these lines are in one plane (5. 2. Sup.). But AB is in the plane in which are BD, DA, because any three straight lines, which meet one another, are in one plane (2. 2. Sup.): Therefore AB, BD, DC are in one plane; and each of the angles ABD, BDC is a right angle ; therefore AB is parallel (28. 1.) to CD. Wherefore, if two straight lines, &c. Q. E. D. PROP. VII., THEOR. If two straight lines be parallel, and one of them at right angles to os plane; the other is also at right angles to the same plane. Let AB, CD be two parallel straight lines, and let one of them AB be at right angles to a plane ; the A other CD is at right angles to the same plane. For, if CD be not perpendicular to the plane to which AB is perpendicular, let DG be perpendicular to it. Then (6. 2. Sup.) DG is parallel to AB : DG and DC there. E fore are both parallel to AB, and are drawn through the same point, B D D, which is impossible (11. Ax. 1.). F Therefore, &c. Q. E. D. PROP. VIII. THEOR. Two straight lines which are each of them parallel to the same straight line, though not both 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. 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, it is perpendicalar (4. 2. Sup.) to the plane HGK passing through them : and' EF is parallel to AB ; therefore AB is at right angles (7.62. Sup.) to the plane HGK. For the same reason, CD is A H B В likewise at right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane HGK. But G if two straight lines are at right angles to the same plane, they are parallel (6. 2. Sup.) to one 0 D K another. Therefore AB is parallel to CD. Wherefore two straight lines, &c. Q. E. D. If two straight lines meeting one another be parallel to two others that meet one another, though 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 раrallel to the two straight lines DE, EF that meet one another, and are not in the same plane with AB, BC. · The B 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,therefore AD is (33. 1.) 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. But straight lines that are parallel to the same E straight line, though not in the same plane with it, are parallel (8. 2. Sup.) to one another. Therefore AD is parallel to CF; and it is equal to it, and AC, DF join them towards D F the same parts; and therefore (33. 1.) AC is equal and parallel to DF. And because AB, BC are equal to DE, EF, and the base AC to the base DF; the angle ABC is equal (8. 1.) to the angle DEF. Therefore, if two straight lines, &c. Q. È. D. PROP. X. 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. А а In the plane draw any straight line BC, and from the point A draw (12. 1.) 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 (11. 1.), in the plane BH, the straight line DE at right angles BC ; and from the point A draw H AF perpendicular to DE ; and F through F draw (31. 1.) GH parallel to BC: and because BC is at right angles to ED, and DA, BC is at right angles (4. 2. Sup.) B В D С 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 (7. 2. Sup.) angles to the same plane ; wherefore GH is at right angles to the plane through ED, DA, and is perpendicular (def. 1. 2. Sup.) to every straight line meeting it in that plane. But AF, which is in the plane through ED, DA, meets it: Therefore GH is perpendicular to AF, and consequently AF is perpendicular to GH; and AF is also 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 is also at right angles to the plane passing through them (4. 2. Sup.). And the plane passing through ED, GH is the plane BH ; therefore AF is perpendicular to the plane BH ; so that, from the given point A, above the plane BH, the straight line AF is drawn perpendicular to that plane. Which was to be done. Cor. If it be required from a point C in a plane to erect a per. pendicular to that plane, take a point A above the plane, and draw AF perpendicular to the plane ; then, if from C a line be drawn parallel to AF, it will be the perpendicular required; for being parallel to AF it will be perpendicular to the same plane to which AF is perpendicular (7. 2. Sup.). PROP. XI. THEOR. From the same point in a plane, there cannot be two straight lines at right angles to the plané, upon the same side of it: And there can be 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, AC ; the common section of this plane with the given plane is a straight (3. 2. Sup.) line passing through A : Let DAE be their common section : Therefor the straight lines AB, AC, DAE are in one plane : And because CA is at right angles to the given plane, it makes right angles with every straight line meeting it in that plane. But DAE, which is in that plane, meets B 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 abové a plane, there can be but one per D pendicular to that plane ; for if there E could be two, they would be parallel (6.2. Sup.) to one another, which is absurd. Therefore, from the same point, &c. Q. E. D. V PROP. XII. 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 must meet one another when produced, and their common section must be a straight line GH, in G which take any point K, and join AK, BK : Then, because AB is perpendicular to the plane EF, it is perpendicular (def. 1. 2. Sup.) K 지 to the straight line BK which is in that plane, C and therefore ABK is a right angle. For the HT same reason, BAK is a right angle ; where SE fore the two angles ABK, BAK of the trian A B gle ABK are equal to two right angles, which is impossible (17. 1.): Therefore the planes CD, EF, though produced, do not meet one another ; that is, they are parallel (def. 7. 2. Sup.). Therefore planes, &c. Q. E. D. D PROP. XIII. THEOR. If two straight lines meeting one another, be parallel to two straight lines which also meet one another, but are not in the same plane with the first two: the plane which passes through the first two 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, ĘF shall not meet, though produced. From the point B draw BG perpendicular (10. 2. Sup.) to the plane which passes through DE, EF, and let it meet that plane in G; and through G draw GH parallel to ED (31. 1.), and GK parallel to EF : D And because BG is perpendicular to the plane through DE, EF, jt must make right angles with every straight line meeting E it in that plane (1. def. 2. Sup.). But the straight lines B G GH, GK in that plane meet it: Therefore each of the C K angles BGH, BGK is a right angle : And because BA is parallel (8. 2. Sup.) to GH (for each of them is paral Α. I lel to DE), the angles GBA, BGH are together equal (29. 1.) to two right angles : And BGH is a 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 (4. 2. Sup.) to the plane through BA, BC: And it is 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 (12. 2. Sup.) to one another: Therefore the plane through AB, BC, is, parallel to the plane through DE, EF. Wherefore, if two straight lines, &c. Q. E. D. Cor. It follows from this demonstration, that if a straight line meet two parallel planes, and be perpendicular to one of them, it must be perpendicular to the other also. PROP. XIV. THEOR. If two parallel planes be cut by another plane, their common sectious with it are parallels. Let the parallel planes AB, CD be cut by the plane EFHG, and let their common sections with it be EF, GH ; EF is paral H. lel to GH. For the straight lines EF B and GH are in the same plane, viz. EFHG, which cuts the planes AB and.CD; and they do not meet though produced; for the planes in which they are do not meet ; therefore A EF and GH are parallel (def. 30. 1.). Q. E. D. |