THEOREM. 319. One part of a straight line cannot be in a plane, and another part without it. Demonstration. By the definition of a piane (6) a straight line, which has two points in common with the plane, lies wholly in that plane. 320. Scholium. In order to determine whether a surface is plane, it is necessary to apply a straight line in different directions to this surface, and see if it touches the surface in its whole extent. Fig. 181. THEOREM. 321. Two straight lines which cut each other are in the same plane, and determine its position. Demonstration. Let AB, AC (fig. 181), be two straight lines which cut each other in A. Conceive a plane to pass through AB, and to be turned about AB, until it passes through the point C; then, two points A and C being in the plane, the whole line AC is in this plane; therefore the position of the plane is determined by the condition of its containing the two lines AB, AC. 322. Corollary 1. A triangle ABC, or three points A, B, C, not in the same straight line, determine the position of a plane. Fig. 182. 323. Corollary II. Also two parallels AB, CD (fig. 182), determine the position of a plane; for, if the line EF be drawn, the plane of the two straight lines AE, EF, will be that of the parallels AB, CD. Fig 183. THEOREM. 324. If two planes cut each other, their common intersection is a straight line. Demonstration. If, among the points common to the two planes, there were three not in the same straight line, the two planes in question, passing each through these three points, would make only one and the same plane, which is contrary to the supposition. THEOREM. 325. If a straight line AP (fig. 183) is perpendicular to two others PB, PC, which intersect each other at its foot in the plane MN, it will be perpendicular to every other straight line PQ drawn through its foot in the same plane, and thus it will be perpendicular to the plane MN. Demonstration. Through a point Q, taken at pleasure in PQ, draw the straight line BC in the angle BPC, making BQ = QC (242); join AB, AQ, AC. The base BC being bisected at the point Q, the triangle BPC will give PC+PB=2PQ+2QC (194). The triangle BAC will give, in like manner, AC +AB=2AQ+2QČ. If we subtract the first equation from the second, and recollect that the triangles APC, APB, each right-angled at P, give AČ—PC — AP, AB — PB = AP, we shall have = -2 AP +AP = 2AQ—2PQ; or, by taking half of each member, AP = AQ — PQ; bence AP+PQ=AQ; therefore the triangle APQ is right-angled at P (193), and AP is perpendicular to PQ. 326. Scholium. It is evident, then, not only that a straight line may be perpendicular to all those which pass through its foot in the plane, but that this happens whenever the line in question is perpendicular to two straight lines drawn in the plane; hence the propriety of the definition, art. 313. 327. Corollary 1. The perpendicular AP is less than any oblique line AQ; therefore it measures the true distance of a point A from the plane PQ. 328. Corollary 11. Through any given point P in a plane, only one perpendicular can be drawn to this plane; for, if there could be two, a plane being supposed to pass through them, intersecting the plane MN in PQ, the two perpendiculars would be perpendicular to the line PQ at the same point, and in the same plane, which is impossible (50). It is also impossible to let fall from a given point, without a plane, two perpendiculars to this plane; for, let AP, AQ, be these two perpendiculars, then the triangle APQ would have two right angles APQ, AQP, which is impossible. Fig. 184. Fig. 185. THEOREM. 329. Oblique lines equally distant from the perpendicular are equal; and of two oblique lines unequally distant from the perpendicular, that which is at the greater distance is the greater. Demonstration. The angles APB, APC, APD (fig. 184), being right angles, if we suppose the distances PB, PC, PD, equal to each other, the triangles APB, APC, APD, have two sides and the included angle respectively equal; they are consequently equal; therefore the hypothenuses, or the oblique lines AB, AC, AD, are equal to each other. Likewise, if the distance PE is greater than PD or its equal PB, it is evident that the oblique line AE will be greater than AB or its equal AD. 330. Corollary. All the equal oblique lines AB, AC, AD, &c., terminate in the circumference of a circle BCD described about the foot of the perpendicular P, as a centre; therefore, a point A without a plane being given, to find the point P where the perpendicular A meets this plane, take three points B, C, D, equally distant from the point A, and find the centre of the circle which passes through these points; this centre will be the point P required. 331. Scholium. The angle ABP is called the inclination of the oblique line AB to the plane MN. It is manifest that this inclination is the same for all the oblique lines AB, AC, AD, &c., which depart equally from the perpendicular; for all the triangles ABP, ACP, ADP, &c., are equal. THEOREM. 332. Let AP (fig. 185) be a perpendicular to the plane MN, and BC a line situated in this plane; if, from the foot P of the perpendicular, a line PD be drawn perpendicular to BC, and AD be joined, AD will be perpendicular to BC. Demonstration. Take DB = DC, and join PB, PC, AB, AC. Since DBDC, the oblique line PB = PC; and, because PB PC, the oblique lines AB, AC, considered with reference to the perpendicular AP, are equal (329); hence the line AD has two points A and D each equally distant from the extremities B and C; therefore AD is perpendicular to BC (55). 333 Corollary. It will be seen, at the same time, that BC is perpendicular to the plane APD, since BC is perpendicular at the same time to the two straight lines AD and PD. 334. Scholium. The two lines AE, BC, present an example of two lines which do not meet, because they are not situated in the same plane. The least distance of these lines is the straight line PD, which is at the same time perpendicular to the line AP and to the line BC. The distance PD is the shortest; because, if we join two other points, as A and B, we shall have AB>AD, AD>PD, and, for a still stronger reason, AB>PD. The two lines AE, CB, although not situated in the same plane, are considered as making a right angle with each other, because AE and a line drawn through any point in it parallel to BC, would make a right angle with each other. In like manner, the line AB and the line PD, which represent two straight lines not situated in the same plane, are considered as making the same angle with each other, as is made by AB and a line parallel to PD drawn through some point in AB. THEOREM. 335. If the line AP (fig. 186) is perpendicular to the plane Fig. 186. MN, every line DE parallel to AP will be perpendicular to the same plane. Demonstration. Let there be a plane passing through the parallels AP, DE, intersecting the plane MN in PD; in the plane MN draw BC perpendicular to PD, and join AD. According to the corollary of the preceding theorem, BC is perpendicular to the plane APDE; consequently the angle BDE is a right angle; but the angle EDP is also a right angle, since AP is perpendicular to PD, and DE is parallel to AP (73); hence the line DE is perpendicular to each of the lines DP, DB; therefore it is perpendicular to the plane MN passing through them (325). 336. Corollary 1. Conversely, if the straight lines AP, DE, are perpendicular to the same plane MN, they will be parallel; for, if they are not, through the point D draw a line parallel to AP; this parallel will be perpendicular to the plane MN; consequently there would be two perpendiculars to the same plane drawn through the same point, which is impossible (328). 337. Corollary 11. Two lines A and B, parallel to a third C, are parallel to each other; for, let there be a plane perpendicular to the line C, the lines A and B parallel to this perpendicu Fig. 187. Fig. 188. lar will be perpendicular to the same plane; therefore, by the above corollary, they are parallel to each other. It is supposed that the three lines are not in the same plane, without which the proposition would already be known (77). THEOREM. 338. If the straight line AB (fig. 187) is parallel to another straight line CD, drawn in the plane MN, it will be parallel to this plane. Demonstration. If the line AB, which is in the plane ABCD, should meet the plane MN, this can take place only in some point of the line CD, the common intersection of the two planes; now AB cannot meet CD, because it is parallel to it; consequently it cannot meet the plane MN; therefore it is parallel to this plane (314). THEOREM. 339. Two planes MN, PQ (fig. 188), perpendicular to the same straight line AB, are parallel to each other. Demonstration. If they can meet, let O be one of the common points of intersection, and join OA, OB; the line AB, perpendicular to the plane MN, is perpendicular to the straight line OA drawn through its foot in this plane; for the same reason, AB is perpendicular to BO; hence OA, OB, would be two perpendiculars let fall from the same point O upon the same straight line, which is impossible; consequently the planes MN, PQ, cannot meet; therefore they are parallel. Fig. 189. Fig. 188. THEOREM. 340. The intersections EF, GH (fig. 189), of two parallel planes MN, PQ, by a third plane FG, are parallel. Demonstration. If the lines EF, GH, situated in the same plane, are not parallel, being produced they will meet; consequently the planes MN, PQ, in which they are, would meet; therefore they would not be parallel. THEOREM. 341. The line AB (fig. 188), perpendicular to the plane MN, is perpendicular to the plane PQ, parallel to the plane MN. |