Demonstration. In the plane PQ draw at pleasure the line BC, and through AB, BC, suppose a plane ABC to pass intersecting the plane MN in AD, the intersection AD will be parallel to BC (340); but the line AB, perpendicular to the plane MN, is perpendicular to the straight line AD; consequently it will be perpendicular to its parallel BC; and, since the line AB is perpendicular to every line BC drawn through the foot of it in the plane PQ, it follows that it is perpendicular to the plane PQ. THEOREM. 342. The parallels EG, FH (fig. 189), comprehended between Fig 189 two parallel planes MN, PQ, are equal. Demonstration. Through the parallels EG, FH, suppose a plane EGHF to pass meeting the parallel planes in EF, GH. The intersections EF, GH, are parallel (340) as well as EG, FH; consequently the figure EGHF is a parallelogram; therefore EG FH. = 343. Corollary. It follows from this, that two parallel planes are throughout at the same distance from each other; for, if EG, FH, are perpendicular to the two planes MN, PQ, they are parallel to each other (335); therefore they are equal. THEOREM. 344. If two angles CAE, DBF (fig. 190), not in the same plane, Fig. 190. have their sides parallel, and directed the same way, these angles will be equal, and their planes will be parallel. Demonstration. Take AC BD, AE = BF, and join CE, DF, AB, CD, EF. Since AC is equal and parallel to BD, the figure ABDC, is a parallelogram (84); therefore CD is equal and parallel to AB. For a similar reason, EF is equal and parallel to AB; consequently CD is also equal and parallel to EF, hence the figure CEFD is a parallelogram, and thus the side CE is equal and parallel to DF; the triangles, then, CAE, DBF, are equilateral with respect to each other; therefore the angle CAE = DBF. Again, the plane ACE is parallel to the plane BDF; for, let us suppose the plane parallel to DBF, drawn through the point A, to meet the lines CD, EF, in points different from C and E, for example, in G and H; then, according to article 342, the GEOM. 15 three lines AB, GD, FH, will be equal; but the three AB, CD, EF, are also equal; hence we should have CD GD, and FH FE, which is absurd; therefore the plane ACE is parallel to BDF. 345. Corollary. If two parallel planes MN, PQ, are met by two other planes CABD, EABF, the angles CAE, DBF, formed by the intersections in the parallel planes, are equal; for the intersection AC is parallel to BD (340), and AE to BF; therefore the angle CAE = DBF. THEOREM. 346. If three straight lines not in the same plane AB, CD, EF Fig. 190. (fig. 190), are equal and parallel, the triangles ACE, BDF, formed by joining the extremities of these lines, on the one hand and on the other, will be equal and their planes will be parallel. Demonstration. Since AB is equal and parallel to CD, the figure ABDC is a parallelogram; consequently the side AC is equal and parallel to BD. For a similar reason the sides AE, BF, are equal and parallel, as also CE, DF; hence the two triangles ACE, BDF, are equal; it may be shown moreover, as in the preceding proposition, that their planes are parallel. Fig 191. THEOREM. 347. Two straight lines comprehended between three parallel planes are divided into parts that are proportional to each other. Demonstration. Let us suppose that the line AB (fig. 191) meets the parallel planes MN, PQ, RS, in A, E, B, and that the line CD meets the same planes in C, F, D, we shall have AE: EB:: CF: FD. Draw AD meeting the plane PQ in G, and join AC, EG, GF, BD; the intersections EG, BD, of the parallel planes PQ, RS, by the plane ABD, are parallel (340); hence, AE: EB::AG: GD; and, because the intersections AC, GF, are parallel, AG: GD: CF: FD; therefore, on account of the common ratio, AG: GD, we have AE: EB:: CF: FD. THEOREM. Fig. 192. 348. Let ABCD (fig. 192) be any quadrilateral either in the same plane or not; if the opposite sides are cut proportionally by two straight lines EF, GH, so that AE: EB:: DF: FC, and BG : GC :: AH: HD, the straight lines EF, GH, will cut each other in a point M, in such a manner that HM: MG :: AE: EB, • EM: MF :: AH : HD. Demonstration. Let there be any plane Ab Hc D passing through AD which does not pass through GH; through the points E, B, C, F, draw Ee, Bb, C c, Ff, parallel to GH meeting this plane in è, b, c, f. On account of the parallels Bb, GH, Cc, bH: Hc :: BG: GC:: AH: HD (196); consequently the triangles AHb, DH c, are similar (208). Also and hence but, on account of the similar triangles AHb, DHc, = Besides, the triangles AHb, DHc, being similar, the angle IIA e = HDƒ; hence the triangles AHe, DHƒ, are similar (208), and consequently the angle AHe DHf. It follows then, in the first place, that e Hƒ is a straight line, and that thus the three parallels Ee, GH, Fƒ, are situated in the same plane which contains the two straight lines EF, GH; therefore these must cut each other in a point M. Moreover, on account of the parallels Ee, MH, Ff, EM: MF::e H: Hƒ:: AH: HD. By a similar construction, referred to the side AB, it may be demonstrated that HM: MG :: AE : EB. THEOREM. 349. The angle comprehended between two planes MAN, MAP, may be measured, conformably to the definition, by the angle PAN (fig. 193) made by the two lines AN, AP, drawn one in one Fig. 195. of these planes, and the other in the other perpendicularly to the common intersection AM. Demonstration. In order to show the legitimacy of this measure it is necessary to prove, 1. that it is constant, or, in other words, that it is the same to whatever point of the common intersection the two perpendiculars are drawn. If we take another point M, and draw MC in the plane MN, and MB in the plane MP, perpendicular to the common intersec tion AM; since MB and AP are perpendicular to the same line AM, they are parallel to each other. For the same reason MC is parallel to AN; consequently the angle BMC = PAN (344); therefore, whether the perpendiculars be drawn to the point M or to the point A, the angle is always the same. 2. It is necessary to show that, if the angle of the two planes increases or diminishes, the angle PAN increases and diminishes in the same ratio. In the plane PAN describe, from the centre A, and with any radius, the arc NDP, and from the centre M, and with the same radius, the arc CEB; draw AD to any point D in the arc NP; the two planes PAN, BMC, being perpendicular to the same straight line MA, are parallel to each other (339); consequently the intersections AD, ME, of the two planes by the third AMD, are parallel; therefore the angle BME is equal to PAD (344). Calling, for the present, the angle formed by the two planes MP, MN, a wedge, if the angle DAP were equal to DAN, it is evident that the wedge DAMP would be equal to the wedge DAMN; for the base PAD might be applied exactly to its equal DAN, and the altitude AM would be the same for both; therefore the two wedges would coincide with each other. It is manifest, likewise, if the angle DAP were contained a certain number of times without a fraction in the angle PAN, the wedge DAMP would be contained as many times in the wedge PAMN. Moreover, from a ratio in an entire number to any ratio whatever the conclusion is legitimate, and has been demonstrated in a case altogether similar (122); consequently, whatever be the ratio of the angle DAP to the angle PAN, the wedge DAMP will have the same ratio to the wedge PAMN; therefore the angle NAP may be taken for the measure of the wedge PAMN, or of the angle made by the two planes MAP, MAN. 350. Scholium. It is with angles formed by two planes as it is with angles formed by two straight lines. Thus, when two planes intersect each other, the angles opposite to each other at the vertex are equal, and the adjacent angles are together equal to two right angles; therefore, when one plane is perpendicular to another, the latter is perpendicular to the former. Also, when two parallel planes are intersected by a third plane, the same properties exist with respect to the angles thus formed, as take place, when two parallel lines are met by a third line (73). THEOREM. 351. The line AP (fig. 194) being perpendicular to the plane Fig. 194. MN, any plane APB, passing through AP, will be perpendicular to the plane MN. Demonstration. Let BC be the intersection of the planes AB, MN; if, in the plane MN, the line DE be drawn perpendicular to BP, the line AP, being perpendicular to the plane MN, will be perpendicular to each of the two straight lines BC, DE. But the angle APD formed by the two perpendiculars PA, PD, at the common intersection BP, measures the angle of the two planes AB, MN; therefore, since this angle is a right angle, the two planes are perpendicular to each other (317). 352. Scholium. When three straight lines, as AP, BP, DP, are perpendicular to each other, each of these lines is perpendicular to the plane of the two others, and the three planes are perpendicular to each other. THEOREM. 353. If the plane AB (fig. 194) is perpendicular to the plane Fig. 194 MN, and in the plane AB the line AP be drawn perpendicular to the common intersection PB, the line AP will be perpendicular to the plane MN. Demonstration. If, in the plane MN, the line PD be drawn perpendicular to PB, the angle APD will be a right angle, since the planes are perpendicular to each other; consequently, the line AP is perpendicular to the two straight lines PB, PD; therefore it is perpendicular to the plane MN. 354. Corollary. If the plane AB is perpendicular to the plane MN, and if through a point P of the common intersection a perpendicular to the plane MN be drawn, this perpendicular will be in the plane AB; for, if it is not, there may be drawn, in the plane AB, a line AP perpendicular to the common intersection BP, which would be at the same time perpendicular to the plane MN; therefore there would be two perpendiculars to the plane MN at the same point P, which is impossible (325). THEOREM. 355. If two planes AB, AD (fig. 194), are perpendicular to a Fig. 194. third MN, their common intersection AP will be perpendicular to this third plane. |