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 p'ane BH, the straight line A E DE at right angles to BC; and 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 stand 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 endicular 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 AF it will be perpendicular to the same plane to which AF is per pendicular (7. 2. Sup.). PROP. XI. THEOR. to From the same point in a 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 it. 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: Therefore 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 meetingit 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 B C above a plane, there can be but one per- D pendicular to that plane; for if there 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. 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. K If not, they must meet one another when produced, and their common section must 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 (def. 1. 2. Sup.) to the straight line BK which is in that plane, C and 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 (def. 7. 2. Sup.). Therefore planes, &c. Q. E. D. PROP. XIII. THEOR. A D H F B E 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, EF 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: And because BG is perpendicular to the plane through DE, EF, it must make right angles with E every straight line meeting it in that plane (1. def. 2. Sup.). 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. 2. Sup.) to GH DH 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 sections with it are parallels. Let the parallel planes AB, CD be cut by the plane EFHG, and let their common sections with A PROP. XV. THEOR. If 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 section of the plane EH with the two planes AB 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 the straight lines KM, KN, let a plane be made to pass, cutting the plane CD in the line LO. And because EF and GH are the common sections of the plane EH with the two_parallel planes AB and CD, EF is parallel to GH (14. 2. Sup.). But EF is at right angles to the plane that passes through KN and KM (4. 2. Sup.), because it is at right angles to the lines KM and KN: therefore GH is also at right angles to the same plane (7. 2. Sup.), and it is therefore at right angles to the lines LM, LO which it meets in that plane. Terefore, since LM and LO are at right angles to LG, the common section of the two planes CD and EH. the angle OLM is the inclination of the plane CD to the plane EH (4. def. 2. Sup.). For the same reason the angle MKN is the inclination of the plane AB to the plane EH. But Because KN and LO are parallel, being the common sections of the parallel planes AB and CD with a third plane, the interior angle NKM is equal to the exterior angle OLM (29. 1.); that is, the inclination of the plane AB to the plane EH, is equal to the inclination of the plane CD to the same plane EH. Therefore, &c. Q. E. D. PROP. XVI. THEOR. If two straight lines be cut by parallel planes, they must 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 H parallel planes KL, MN are cut by lines, &c. Q. E. D. PROP. XVII. THEOR. If a straight line be at right angles to a plane, every plane whick passes through that line is at right angles to the first mentioned plane. Let the straight line AB be at right angles to a plane CK; every plane which passes through AB is at right angles to the plane CK. Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK; take any point Fin CE, from which draw FG in the plane DE at right angles to CE: And because AB is perpendicular to the plane CK, therefore it is also perpendicular to every straight line meeting it in that plane (1. def. 2. Sup.); and consequently it is perpendicular to CE: Wherefore ABF is a right angle; But GFB is likewise a right angle; therefore AB is parallel (28. 1.) to FG. And AB is at right angles to the plane CK: therefore FG is |