Demonstration. Because AB is perpendicular to the plane, it is perpendicular to BE and BD (XI. Def. 3); therefore 2. Each of the angles ABD and ABE is a right angle. Because the straight line BD meets the parallel straight lines AB and CD; therefore 3. The angles ABD and CDB are together equal to two right angles (I. 29). But ABD is a right angle; therefore 4. CDB is also a right angle, and CD is perpendicular to BD. But it may be proved, as in the sixth proposition, that ED is perpendicular to DA, and it is also perpendicular to BD (constr.) Therefore ED is perpendicular to the plane of BD and DA (XI. 4); and therefore 5. ED makes right angles with every straight line meeting it in that plane (XI. Def. 3). But DC is in the plane of BD and DA, because all three are in the plane of the parallels AB and CD; therefore 6. ED is at right angles to DC, and CD to DE. But CD is also at right angles to DB; therefore wherefore 7. CD is at right angles to the two straight lines DE and DB at the point of their intersection D; E 8. CD is at right angles to the plane of DE and DB, that is the same plane to which AB is at right angles (XI. 4). Therefore, if two straight lines, &c. Q.E.D. PROPOSITION 9.-Theorem. Two straight lines which are each of them parallel to the same straight line, but not in the same plane with it, are parallel to one another. H K Let AB and CD be each of them parallel to EF, but not in the same plane with it. Then AB is parallel to CD. Construction. In EF, take any point G, and from it, draw, in the plane of EF and AB, the straight line GH at right angles to EF (I. 11); and in the plane of EF and CD, GK at right angles to EF. E H Demonstration. Because EF is perpendicular both to GH and GK; therefore B D 1. EF is perpendicular to the plane HGK passing through them (XI. 4). But EF is parallel to AB (hyp.), therefore For the same reason, Therefore 2. AB is at right angles to the plane HGK (XI. 8). 3. CD is at right angles to the plane HGK. 4. AB and CD are each of them at right angles to the plane But if two straight lines are at right angles to the same plane, they are parallel (XI. 6); therefore 5. AB is parallel to CD. Wherefore, two straight lines, &c. Q.E.D. PROPOSITION 10.-Theorem. If two straight lines meeting one another be parallel to two others, which meet one another, but are not in the same plane with the first two; the first two and the other two contain equal angles. Let the two straight lines AB and BC, meeting one another, be parallel to the two straight lines DE and EF, which meet one another, and are not in the same plane with AB and BC. Then the angle ABC is equal to the angle DEF. Construction. Make BA, BC, ED, and EF all equal to one another. Join AD, CF .BE .AC and DF. fore Demonstration. Because BA is equal and parallel to ED, there 1. AD is equal and parallel to BE (I. 33). For the same reason, CF is equal and parallel to BE; therefore 2. AD and CF are each equal and parallel to BE; but straight lines that are parallel to the same straight line, but not in the same plane with it, are parallel to one another (XI. 9), therefore 3. AD is parallel to CF. Because AD is equal and parallel to CF (I. Ax. 1), and AC and DF join them towards the same parts; therefore 4. AC is equal and parallel to DF (I. 33) ; and because AB and BC are equal to DE and EF, each to each, and the base AC to the base DF; therefore 5. The angle ABC is equal to the angle DEF (I. 8). Wherefore, if two straight lines, &c. Q.E.D. PROPOSITION 11.-Problem. 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. Construction. In the plane BH draw any straight line BC, and in the plane passing through BC and A, from the point A draw AD perpendicular to BC (I. 12). If AD be also perpendicular to the plane BH, what was required is done. But if it be not, from the point D, draw in the plane BH, the straight line DE at right angles to BC (I. 11) and in the plane passing through DE and A, from the point ▲, draw AF perpendicular to DE. Then AF is perpendicular to the plane BH. Through F, draw GH parallel to BC (I. 31). E F D B 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; but 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). D A 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 plane, 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. |