A C E B G K D at right angles to the same EF. And because EF is perpendicular both to GH and GK, EF is perpendicular (4. Pl.) to the plane HGK passing through them: and EF is parallel to AB; therefore A B is at right angles (7. Pl.) 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 HGK. But if two straight lines are at right angles to the same plane, they are parallel (6. Pl.) to one another: therefore AB is parallel to CD. two straight lines, &c. Q. E. D. PROP. IX. THEOR. Wherefore If two straight lines meeting one another be parallel to two others that meet one another, though not in the same plane with the other 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. A B E C Take BA, BC, ED, EF all equal to one another, and draw AD, CF, BE, AC, DF. Because BA is equal and parallel to ED, therefore AD is (Euc. 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 straight line, though not in the same plane with it, are parallel (8. Pl.) to one another: therefore AD is parallel to CF; and it is equal to it, and AC, DF join them towards the same parts; and therefore (Euc. 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 (Euc. 8. 1.) to the angle DEF. Therefore, if two straight lines, &c. Q. E. D. PROP. X. PROB. D F 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 (Euc. 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 (Euc. 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 (Euc. 31. 1.) GH parallel to BC; and because BC is at right angles to ED and DA, BC is at right angles (4. Pl.) to the plane passing through ED, DA. And G H is parallel to BC; but A B G D P if two straight lines be parallel, one of which is at right angles to a plane, the other shall be at right (7. Pl.) angles to the same plane; wherefore GH is at right angles to the plane passing through ED, DA, and is perpendicular (1 Def. Pl.) 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 shall also be at right angles to the plane passing through them (4. Pl.). 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 вH, the straight line AF is drawn perpendicular to that plane. Q. E. F. COR. If it be required from a point c in a plane to erect a perpendicular 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. Pl.). 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. B C 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. Pl.) 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 be parallel (6. Pl.) to one another, which is absurd. Therefore, from the same point, &c. Q. E. D. PROP. XII. THEOR. D 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. C K 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 A, K, and B, K. Then, because A B is perpendicular to the plane EF, it is perpendicular (1 Def. Pl.) to the 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 (Euc. 17. 1.): therefore the planes CD, EF, though produced, do not meet one another: that is, they are parallel (7. Def. Pl.). Therefore, planes, &c. Q. E. D. PROP. XIII. THEOR. A D B E IF 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 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. E From the point B draw BG perpendicular (10. Pl.) to the plane which passes through DE, EF, and let it meet that plane in G, and through G draw GH parallel to ED (Euc.31.1.), 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 (1. Def. Pl.). 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 (8. Pl.) to GH (for each B A F K of them is a parallel to DE), the angles GBA, BGH are together equal (Euc. 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. Pl.) 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. Pl.) 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. PROP. XIV. THEOR. If two parallel planes be cut by another plane, their common sections with it are parallel. Let the parallel planes AB, CD be cut by the plane EF HG, and let their common sections with it be EF, GH: EF is If a straight line be perpendicular to one of two parallel planes it is perpendicular to the other. Let the plane CD be parallel to the plane AB, and the straight line EF perpendicular to the plane AB: EF is also perpendicular to the plane CD. D -L E K C Having made any two planes pass through E F, let one of them cut the parallel planes in the straight lines FG and EK, and let the other cut the same planes in the straight lines F H and EL. The plane CD, being parallel to the plane AB, EK (14. Pl.), is parallel, to FG, and EL to FH. And because EF is perpendicular to the plane AB, it is perpendicular to FG and to FH (1. Def. Pl.). But a straight line which is perpendicular to one of two parallel straight lines is perpendicular to the other: wherefore EF is perpendicular to each of the straight A F G B H lines EK and EL; and consequently, it is perpendicular to the plane CD, in which EK and EL are situated (4. Pl.). PROP. XVI. THEOR. Ir two parallel planes be cut by a third plane: they have the same inclination to that plane. Let AB and CD be two parallel planes, and EH a third plane cutting them: the planes AB and CD are equally inclined to EH. Let the straight lines EF and GH be the common sections of the plane EH with the two planes A B and CD; and from K, any point in EF, draw in the plane EH the straight line KM at right angles to EF, and let it meet GH in L; draw also KN at right angles to EF in the plane AB; and through |