three lines AB, GD, FH, will be equal; but the three AB, CD, EF, are also equal; hence we should have CD = GD, and FHFE, 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 other, will be equal and their planes will be parallel.

Fig. 191.

Fig. 192.

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.

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 AEEB:: CF: FD.

THEOREM.

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, Cc, Ff, parallel to GH meeting this plane in e, b, c, f. On account of the parallels B b, GH, Cc, bH: Hc:: BG: GC:: AH: HD (196);

consequently the triangles AHb, DHc, are similar (208). Also Ae: eb: AE: EB

but, on account of the similar triangles AH b, DH e,

Besides, the triangles AHb, DH c, being similar, the angle
HA e = HDf; hence the triangles AH, DHƒ, are similar
(208), and consequently the angle AH e = DHf. It follows then,
in the first place, that e Hf is a straight line, and that thus the
three parallels E e, 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
E e, 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 of these Fig. 193. 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 (64).

THEOREM.

351. The line AP (fig. 194) being perpendicular to the plane MN, Fig. 194. 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 MN, Fig. 194. 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.

Demonstration. If through the point P a perpendicular to the plane MN be drawn, this perpendicular must be at the same time in the plane AB and in the plane AD (354); therefore it is their common intersection AP.

THEOREM.

356. If a solid angle is formed by three plane angles, the sum of either two of these angles will be greater than the third.

Demonstration.. We need consider only the case in which the plane angle to be compared with the two others is greater than Fig. 195. either of them. Let there be then the solid angle S (fig. 195)

Fig. 196.

formed by the three plane angles ASB, ASC, BSC, and let us suppose that the angle ASB is the greatest of the three; we say that ASB ASC + BSC.

In the plane ASB make the angle BSD BSC, draw at pleasure the straight line ADB; and, having taken SC=SD, join AC, BC.

=

The two sides BS, SD, are equal to the two BS, SC, and the angle BSD = BSC; hence the two triangles BSD, BSC, are equal; consequently BD BC. But AB < AC+ BC; if we take from the one BD, and from the other its equal BC, there will remain AD <AC. The two sides AS, SD, are equal to the two AS, SC, and the third AD is less than the third AC; therefore the angle ASD <ASC (42). Adding BSD = BSC,we shall have ASD + BSD or ASB < ASC + BSC.

THEOREM.

357. The sum of the plane angles which form a solid angle is always less than four right angles.

Demonstration. Suppose the solid angle S (fig. 196) to be cut by a plane ABCDE; from a point O taken in this plane draw to the several angles the lines OA, OB, OC, OD, OE.

The sum of the angles of the triangles ASB, BSC, &c., formed about the vertex S, is equal to the sum of the angles of an equal number of triangles AOB, BOC, &c., formed about the vertex O. But, at the point B, the angles ABO, OBC, taken together, make the angle ABC less than the sum of the angles ABS, SBC (356); likewise, at the point C, BCO+OCD < BCS + SCD, and so on through all the angles of the polygon ABCDE. It follows then, that of the triangles whose vertex is in O the sum of the angles