Then AF is perpendicular to the plane BH. Through F, draw GH parallel to BC (I. 31). Demonstration. Because BC is at right angles to ED and DA (constr.), therefore 1. BC is at right angles to the plane of ED and DA (XI. 4). But GH is parallel to BC; and if two straight lines be parallel, one of which is at right angles to a plane, the other is also at right angles to the plane (XI. 8); therefore 2. GH is at right angles to the plane of ED and DA; and is perpendicular to every straight line meeting it in that plane (XI. Def. 3). But AF, which is in the plane of ED and DA, meets it; therefore 3. GH is perpendicular to AF, and AF to GH; but AF is perpendicular to DE (constr.), therefore 4. AF is perpendicular to each of the straight lines GH and DE; þut if a straight line stand at right angles to each of two straight lines at the point of their intersection, it is also at right angles to the plane in which they are, that is, the plane BH (XI. 4); therefore 5. AF is perpendicular to the plane BH. Wherefore, from the given point A, above the plane BH, the straight line AF is drawn perpendicular to it. Q.E.F. PROPOSITION 12.- Problem. To draw a straight line at right angles to a given plane, from a point given in the plane. Let A be the given point in the plane. It is required to draw a straight line from the point A at right angles to the plane. Construction. From any point B above the plane, draw BC perpendicular to it (XI. 11), and from A draw AD parallel to BC (I. 31). B Then AD is at right angles to the plane. Demonstration. Because AD and CB are two parallel straight lines, and one of them BC is at right angles to the given plane, therefore 1. The other AD is also at right angles to it (XI. 8). Therefore a straight line AD has been drawn at right angles to a given plane, from a point A given in it. Q.E.F. PROPOSITION 13.-Theorem. From the same point in a given plane, there cannot be two straight lines drawn perpendicular to the plane, upon the same side of it ; and there can be but one perpendicular to a plāne, from a point above it. For, if it be possible, let the two straight lines AB and AC be at right angles to a given plane from the same point A in the plane, and upon the same side of it. Construction. Let a plane pass through BA and AC; and let the straight line DE be their common section (XI. 3). The straight lines AB, AC, and DE are in one plane. Demonstration. Because CA is at right angles to the given plane, therefore 1. CA makes right angles with every straight line meeting it in that plane (XI. Def. 3) ; but DE, which is in that plane, meets CA; therefore 2. CAE is a right angle. For the same reason, BAE is a right angle; wherefore the angle CAE is equal to the angle BAE (I. Ax. 11), therefore 3. CAE and BAE cre in one plane, which is impossible (I. Ax. 9). Therefore two perpendiculars cannot be drawn from the same point in a plane, on the same side of it. Also, from a point above a plane, there can be but one perpendicular to it. For, if there could be two, they would be parallel to one another (XI. 6), which is absurd. Therefore, from the same point, &c. Q.E.D. PROPOSITION 14.-Theorem. 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 and EF. Then the planes CD and EF are parallel to one another. Construction. If not, they shall meet one another when produced. Let them be produced and meet; and let the straight line GH be their common section. In GH, take any point K, and join AK and BK. Demonstration. Because AB is perpendicular to the plane EF, it is perpendicular to the straight line BK meeting it in that plane (XI. Def. 3); therefore 1. ABK is a right angle. For the same reason, BAK is a right angle; therefore 2. The two angles ABK and BAK of the triangle ABK, are equal to two right angles, which is impossible (I. 17); therefore 3. The planes CD and EF, though produced, do not meet one another ; that is, 4. The planes CD and EF are parallel to one another (XI. Def. 8). Therefore, planes to which, &c. Q.E.D. PROPOSITION 15.-Theorem. Two planes are parallel, if two straight lines which meet each other in the one plane, be parallel to two straight lines which meet each other in the other plane. Let AB and BC, two straight lines which meet each other, be parallel to two straight lines DE and EF, which meet each other, but are not in the same plane with AB and BC. Then the plane of AB and BC is parallel to the plane of DE and EF. plane of DE and EF (XI. 11); and let it meet that plane in G. Through G draw GH parallel to ED, and GK parallel to EF (I. 31). Demonstration. Because BG is perpendicular to the plane of DE and EF; therefore 1. BG makes right angles with every straight line meeting it in that plane (XI. Def. 3); but the straight lines GH, GK meet it in that plane ; therefore 2. Each of the angles BGH, BGK is a right angle. And because BA is parallel to GH, for each of them is parallel to DE, and they are not both in the same plane with it (XI. 9), therefore 3. The angles GBA, BGH are together equal to two right angles (I. 29); and BGH is a right angle; therefore also 4. GBA is a right angle, and GB perpendicular to BA; for the same reason, GB is perpendicular to BC. Because the straight line GB stands at right angles to the two straight lines BA and BC meeting one another in B; therefore 5. GB is perpendicular to the plane of BA and BC (XI. 4.); but it is perpendicular to the plane of DĘ and EF (constr.), therefore 6. BG is perpendicular to each of the planes of AB and BC, and of DE and EF; but planes to which the same straight line is perpendicular are parallel to one another (XI. 14); therefore 7. The plane of AB and BC is parallel to the plane of DE and EF. Wherefore, two planes are, &c. Q.E.D. |