therefore AD is (33. 1.) both equal and pa- 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. E. D. 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. A 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 to BC: and from the point A draw AF perpendicular to DE; and 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. 11.) to the plane passing through ED, DA. And GH is parallel to BC; but, if two straight G lines be parallel, one of which is at right angles to a plane, the other shall be at right (8. 11.) angles to the same plane; wherefore GH is at right angles to the plane through ED, DA, B E F H D C and is perpendicular (3. def. 11.) 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 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 life 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 an gles to the plane. A D B C From any point B above the plane draw (11. 11.) BC perpendicular to it; and from A draw (31. 1.) 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 (8. 11.). 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. 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, AC; the common section of this with the given plane is a straight (3. 11.) 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 B C be but one perpendicular to that plane; D for, if there could be two, they would be parallel (6. 11.) to one another, which is, absurd. Therefore, from the same point, &c. Q. E. D. PROP. XIV. THEOR. PLANES to which the same straight line is perpendicu lar, 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 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 (3. def. 11.) to the C straight 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 right angles, which is impossible (17.1.); therefore the planes CD, EF though produced, do not meet one another; that is, they are parallel (8. def. 11.). Therefore, planes, &c. Q. E. D. PROP. XV. THEOR. Ir two straight lines meeting one another, be parallel 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 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. E G F C K From the point B draw BG perpendicular (11. 11.) to the plane which passes through DE, EF, and let it meet that plane in G; and through G draw GH parallel (31. 1.) to ED, and GK parallel to EF; and because BG is perpendicular to the plane through DE, EF, it shall make right angles with every straight line meeting it in that plane (3. def. 11.). B But the straight lines GH, GK in that plane meet it: therefore each of the angles BGH, BGK is a right angle: and because BA is parallel (9. 11.) to GH (for each A of them is parallel to DE, and they are not both in the same D H plane with it) 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. 11.) to the plane through AB, 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 (14. 11.) 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. * See Note. PROP. XVI. THEOR. If two parallel planes be cut by another plane, their common sections 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 parallel to GH. F K B D For, if it be not, EF, GH shall meet, if produced, either on the side of FH, or EG; first, let them be produced on the side of FH, and meet in the point K; therefore, since EFK is in the plane AB, every point in EFK is in that plane; and K is a point in EFK; therefore K is in the plane AB for the same reason K is also in the plane CD: wherefore the planes AB, CD produced meet one another; but they do not meet, since they are parallel by the hypothesis: therefore the straight lines EF, GH do not meet when produced on the side of FH; in the same manner it may be proved, that EF, GH do not meet when produced on the side of EG: but straight lines which are in the same plane and do not meet, though produced either way, are parallel therefore EF is parallel to GH. Wherefore, if two parallel planes, &c. Q. E. D. C E G PROP. XVII. THEOR. If two straight lines be cut by parallel planes, they shall be cut in the same ratio. Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D; as AE is to EB, so is CF to FD. Join AC, BD, AD, and let AD meet the plane KL in the point X and join EX, XF: because the two parallel planes KL, MN are cut by the plane EBDX, the common sections • See Note. |