If two straight lines are cut by three parallel planes, they will be divided proportionally. Let the lines AB and CD be cut by the paralled planes MN, PQ, and RS, in the points A, E, B, and C, F, D; then The plane ACD intersects the parallel planes MN and PQ, in the parallel lines AC and GF: hence, AG : GD :: CF: FD. Combining these proportions (B. II., P. IV.), we have, AE : EB :: CF: FD; which was to be proved. Cor. 1. If two lines are cut by any number of parallel planes they will be divided proportionally. Cor. 2. If any number of lines are cut by three parallel planes, they will be divided proportionally. If a line is perpendicular to a plane, every plane passed through the line will also be perpendicular to that plane. Let AP be perpendicular to the plane MN, and let BF be a plane passed through AP: then will BF be perpendicular to MN. In the plane MN, perpendicular to BC, F tion of BF and MN. pendicular to BC and DP (D. 1); and since AP and DP, in the planes BF and MN, are perpendicular to the intersection of these planes at the same point, the angle which they form is equal to the angle formed by the planes (D. 4); but this angle is a right angle: hence, BF is perpendicular to MN; which was to be proved. Cor. If three lines AP, BP, and DP, are perpendicular to each other at a common point P, each line will be perpendicular to the plane of the other two, and the three planes will be perpendicular to each other. If two planes are perpendicular to each other, a line drawn in one of them, perpendicular to their intersection, wil be perpendicular to the other. Let the planes BF and MN be perpendicular to each other, and let the line AP, drawn in the plane BF, be perpendicular to the intersection BC; then will AP be perpendicular to the plane MN. For, in the plane MN, draw PD perpendicular to BC Then because the planes BF and MN are perpen at P. Cor. If the plane BF is perpendicular to the plane MN, and if at a point P of their intersection, we erect plane MN, that perpendicular will For, if not, draw in the plane BF, a perpendicular to the be in the plane BF PA perpendicular to PC, will be perpendicular to the therefore, at the same point P, there are two perpendiculars to the plane MN; which is impossible (P. IV., C. 2). the common intersection ; AP plane MN, by the theorem ; PROPOSITION XVIII. THEOREM. If two planes cut each other, and are perpendicular to a third plane, their intersection is also perpendicular to that plane. H F A Let the planes BF, DH, be perpendicular to MN : then will their intersection AP be perpendicular to MN. For, at the point P, erect a perpendicular to the plane MN; that perpendicular must be in the plane BF, and also in the plane DH (P. XVII., C.); therefore, it is their common intersection AP: which was to be proved. B M N The sum of any two of the plane angles formed by the edges of a triedral angle, is greater than the third. Let SA, SB, and angle: then will the sum formed by them, as ASC third ASB. SC, be the edges of a triedral of any two of the plane angles and CSB, be greater than the If the plane angle ASB is equal to, or less than, either of the other two, the truth of the proposition is evident. Let us suppose, then, that ASB is greater than either. In the plane ASB, construct the angle BSD equal to BSC; draw AB in that plane, at plea sure; lay off SC equal to SD, and draw AC and CB. The triangles BSD and BSC have B D the side SC equal to SD, A mon, and the included angles BSD and BSC equal, by construction; the triangles are therefore equal in all their parts hence, BD is equal to BC. But, from Proposition VII., Book I., we have, BC + CA> BD + DA. Taking away the equal parts BC and BD, we have, CA > DA; hence (B. I., P. IX., C.), we have, angle ASC angle ASD ; and, adding the equal angles BSC and BSD, or, angle ASC angle CSB > angle ASD + angle DSB ;. angle ASC angle CSB > angle ASB; which was to be proved. PROPOSITION XX. THEOREM. The sum of the plane angles formed by the edges of any polyedral angle, is less than four right angles. Let S be the vertex of any polyedral angle whose edges are SA, SB, SC, SD, and SE; then will the sum of the angles about S be less than four right angles. For, pass a plane cutting the edges in the points A, B, C, D, and E, and the faces in the lines AB, BC, CD, DE, and EA. From any point within the polygon thus formed, as 0, draw the straight lines OA, OB, OC, OD, and OE B D We then have two sets of triangles, one set having a common vertex S, the other having a common vertex O, and both having common bases AB, BC, CD, DE, EA. Now, in the set which has the common vertex S, the sum of all the angles is equal to the sum of all the plane angles formed by the edges of the polyedral angle whose vertex is S, together with the sum of all the angles at the bases: viz., SAB, SBA, SBC, &c.; and the entire sum is equal to twice as many right angles as there are triangles. In the set whose common vertex is 0, the sum of all the angles is equal to the four right angles about 0, together with the interior angles of the polygon, and this sum is equal to twice as many right angles as there are triangles. Since |