Let AC, AD, AE (fig. 21) be straight lines drawn from the point A oblique to the plane MN, of which AC, AD are equally distant from AB, perpendicular to the plane MN, that is, CB being joined and also DB, CB is equal to DB: AC shall be equal to AD. Because CB is equal to DB, and AB is common to the two triangles CBA, DBA, and the angles CBA, DBA are right angles, AC is equal AD (I. 4). Next let AC be nearer to AB than AE is, that is, joining EB, let CB be less than EB, AC is less than AE. For the square of BC is less than the square of BE; therefore the squares of CB and BA are less than the squares of EB and BA; that is (I. 47), the square of CA is less than the square of EA, consequently CA is less than EA. Conversely, if AC is equal to AD, BC is equal to BD. the For since the square of AC is equal to the square of AD, squares of AB, BC are equal to the squares of AB, BD (I. 47). Therefore the square of BC is equal to the square of BD; and consequently BC is equal to BD. Also, if AC is less than AE, BC is less than BE. For the square of AC being less than the square of AE, the squares of AB, BC are less than the squares of AB, BE (I. 47). Therefore the square of BC is less than the square of BE; and consequently BC less than BE. Therefore of straight lines drawn, &c.: which was to be proved. PROP. XV. THEOR. If from any point in a straight line oblique to a plane, a perpendicular be drawn to the plane, and the intersections of the oblique line and the perpendicular with the plane be joined by a straight line, the oblique straight line will form a less angle with the line joining the intersections than with any other line meeting it in the plane. From the point A (fig. 22), in the straight line AB which is oblique to the plane MN, let AC be drawn perpendicular to the plane, and BC be joined; the angle which AB makes with BC shall be less than that which it makes with any other line BD meeting it in the plane MN. For take BD equal to BC, and join AD. Then because AC is perpendicular, and AD oblique to the plane, AC is less than AD (Prop. 9). And because BC is equal to BD and AB common to the two triangles ABC, ABD, and AC is less than AD, the angle ABC is less than the angle ABD (I. 25): which was to be proved. Scholium. It is for this reason that the angle ABC is considered to measure the inclination of the straight line AB to the plane MN. PROP. XVI. THEOR. If from the foot of the perpendicular to a given plane a perpendicular be let fall upon any straight line in the plane, a straight line drawn from any point in the perpendicular to the plane, to the foot of the perpendicular on the straight line, will be perpendicular to that line; and the plane passing through the perpendicular to the given plane and that on the straight line, will be perpendicular to that line. Let AB (fig. 23) be perpendicular to the plane MN, and from B let BE be drawn perpendicular to the straight line CD in the plane, let AE be joined: AE is perpendicular to CD. Take EC and ED equal to each other, and join BC, BD, AC, AD. Then because CE is equal to DE, and BE is common to the two triangles BEC, BED, and at right angles to DC, BC is equal to BD (I. 4). And because the oblique lines AC and AD are equally distant from the perpendicular AB (Def. 5), AC is equal to AD (Prop. 14). And because CE is equal to DE, AE common to the two triangles AEC, AED, and AC equal to AD, the angle AEC is equal to the angle AED (I. 8); and therefore AE is at right angles to DC (I. Def. 10). Also the plane ABE is perpendicular to DC. For DC is at right angles to BE and AE, and therefore it is perpendicular to the plane ABE (Prop. 4). : Wherefore, if from the foot of the perpendicular, &c. which was to be proved. PROP. XVII. THEOR. If two straight lines, be at right angles to the same plane they shall be parallel to one another. Let the straight lines AB, CD (fig. 24), meeting the plane MN in the points B and D, be at right angles to it: AB shall be parallel to CD. Join BD; draw in the plane MN, DE at right angles to BD; and join AD. Then because from B, the foot of the perpendicular to the plane MN, BD is drawn perpendicular to DE in the plane, and AD is drawn, from a point in AB, to D, AD is perpendicular to DE (Prop. 16); and ED is at right angles to AD. But it is also at right angles to each of the two BD, CD at the point D in which the three straight lines BD, AD, CD meet; therefore these three straight lines are all in the same plane (Prop. 12). But AB is in the plane in which are BD, AD (Prop. 2); therefore AB, BD, CD are in one plane: and since each of the angles ABD, CDB is a right angle, AB is parallel to CD (I. 28). Wherefore, if two straight lines, &c. which was to be proved. PROP. XVIII. THEOR. If two straight lines be parallel, and one of them be at right angles to a plane, the other also shall be at right angles to the same plane. per Let AB, CD (fig. 24) be two parallel straight lines meeting the plane MN in B, D, and let one of them, AB, be at right angles to the plane; the other, CD, shall be at right angles to the same plane. Join BD, AD; then AB and CD being parallel, they are in the same plane (I. Def. 35); and BD and DA are also in that plane (I. Def. 7). In the plane MN draw DE at right angles to BD. Then because AB is perpendicular to the plane MN, DE is pendicular to the plane ABD (Prop. 16), that is, to the plane ABDC; consequently CDE is a right angle (Def. 3). And because BD meets the parallels AB, CD, the angles ABD, CDB are together equal to two right angles (I. 29); but ABD is a right angle (Def. 3); therefore CDB is also a right angle. And it has been shown that the angle CDE is a right angle; therefore CD is at right angles to BD, DE at their point of intersection, and is consequently at right angles to the plane BDE, that is, to the plane MN. Therefore, if two straight lines, &c.: which was to be proved. PROP. XIX. THEOR. Two straight lines, which are each of them parallel to the same straight line, and not both in one plane with it, are parallel to one another. Let AB, CD (fig. 25) be each of them parallel to EF, and not both in one plane with it: AB shall be parallel to CD. In EF take any point G, from which draw, in the plane passing through EF, AB, the straight line GH at right angles to EF (I. 11); and in the plane passing through EF, CD, draw GK at right angles to the same EF. Because EF is perpendicular both to GH and GK, EF is perpendicular to the plane HGK passing through them (Prop. 4): and EF is parallel to AB; therefore AB is at right angles to the plane HGK (Prop. 18). 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 to one another (Prop. 17): therefore AB is parallel to CD. Wherefore two straight lines, &c. ; which was to be proved. Cor. If through two parallel straight lines AB, CD, planes which cut each other be drawn, their intersection EF will be parallel to AB and CD. For from any point E, in EF, draw, in the plane ABFE, a straight line parallel to AB; this line will, by the proposition, be parallel to CD, and therefore in the plane DCEF: it must consequently be the intersection of the two planes passing through AB, CD. PROP. XX. THEOR. If two straight lines meeting one another be parallel to two others that meet one another, and are not in the same plane with the first two, the first two and the other two shall contain equal angles. Let the two straight lines AB, BC (fig. 26.), 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 shall be equal to the angle DEF. Take BA, BC, ED, EF all equal to one another; and join AD, CF, BE, AC, DF. Then because BA is equal and parallel to ED, therefore AD is both equal and parallel to BE (I. 33). For the same reason, CF is equal and parallel to BE. Therefore AD and CF are each of them equal and parallel to BE; and they are not both in one plane with it, therefore AD is parallel to CF (Prop. 19); and it is also equal to it (I. Ax. 1); and AC, DF join AD, CF towards the same parts; therefore AC is equal and parallel to DF (I. 33). And because AB, BC are equal to DE, EF, each to each, and the base AC to the base DF; the angle ABC is equal to the angle DEF (I. 8). Therefore, if two straight lines, &c. : which was to be proved. PROP. XXI. THEOR. If a straight line without a plane be parallel to a straight line in the plane, it will be parallel to the plane. Let the straight line CD (fig. 27) without the plane MN be parallel to the straight line AB in that plane; CD will be parallel to the plane MN. For CD being in the same plane as AB (I. Def. 35), if, being produced, it meet the plane MN, it must meet it in the line AB produced; and this is impossible, since CD is parallel to AB. Therefore CD is parallel to the plane MN: which was to be proved. PROP. XXII. THEOR. If a straight line be perpendicular to a plane, any straight line perpendicular to this line will be parallel to the plane. Let the straight line AB be perpendicular to the plane MN, and the straight line AC be perpendicular to AB; AC is parallel to the plane MN. Through AC and AB (fig. 28), let the plane AD be drawn, intersecting the plane MN in BD. Then since AB is perpendicular to the plane MN, ABD is a right angle; and BAC is a right angle by hypothesis: therefore AC is parallel to BD (I. 28), and consequently to the plane MN (Prop. 21) : which was to be proved, PROP. XXIII. THEOR. If a straight line be parallel to a plane, the intersection with this plane of any plane passing through the line will be parallel to the line. Let the straight line CD (fig. 27) be parallel to the plane MN, and let AB be the intersection of a plane passing through CD with the plane MN: AB is parallel to CD. Since CD is parallel to the plane it cannot meet any line AB in the plane. AB being in the same plane with CD is therefore parallel to it (I. Def. 35): which was to be proved. Cor. 1. CD being parallel to the plane MN, if through any point A, in the plane MN, a straight line AB be drawn parallel to CD, it will be wholly in the plane MN. For this line must be the intersection of the plane passing through CD and A, with the plane MN. Cor. 2. A straight line parallel to each of two planes which cut one another is parallel to their intersection. For if through any point of the intersection of the two planes a straight line be drawn parallel to the former line, it will be in each of the two planes, and will therefore be their intersection. Parallel straight lines, contained between a plane and a straight line parallel to it, are equal. Let AC and BD (fig. 27) be parallel straight lines, contained between the plane MN and the straight line CD parallel to MN; AC and BD are equal. For as the plane of the parallels AC, BD intersects the plane MN in a line AB parallel to CD (Prop. 23), ABDC is a parallelogram : consequently AC is equal to BD (I. 34): which was to be proved. PROP. XXV. THEOR. Planes, to which the same straight line is perpendicular, are parallel to one another. Let the straight line AB (fig. 29) be perpendicular to each of the planes CD, EF: these planes shall be parallel to one another. If not, they shall meet one another when produced: let them meet; their common section is 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 to the straight line BK which is in that plane (Def. 3): 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 (I. 17): therefore the planes CD, EF, though produced, do not meet one another; that is (Def. 10), they are parallel. Therefore planes, &c.: which was to be proved. |