Suppose the line AB to meet the parallel planes MN, PQ, RS, at the points A, E, B; and the line CD to meet the same planes at the points C, F, D: T we are now to show that B D A E B PROPOSITION XV. THEOREM. If two straight lines be cut by three parallel planes, they will be divided proportionally. M H G F AE EB: CF : FD. Draw AD meeting the plane PQ in G, and draw AC, EG, GF, BD; the intersections EG, R BD, of the parallel planes PQ, RS, by the plane ABD, are parallel (Prop. X.); therefore AE EB in like manner, the intersections AC, GF, being parallel, AG: GD:: CF : FD ; the ratio AG: GD is the same in both; hence AG: GD; AE EB:: CF : FD. N D N PROPOSITION XVI. THEOREM. If a lis perpendicular to a plane, every plane passed through rpendicular, will also be perpendicular to the plane. the Let AP be perpendicular to the plane NM; then will every plane passing through AP be perpendicular to NM. M Let BC be the intersection of the planes AB, MN; in the plane MN, draw DE perpendicular to BP: then 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, to the common intersection BP, measures the angle of the two planes AB, MN (Def. 4.); therefore, since that angle is a right angle, the two planes are perpendicular to each other. B D M Scholium. When three straight lines, such as AP, BP, DP, are perpendicular to each other, each of those lines is perpendicular to the plane of the other two, and the three planes are perpendicular to each other. PROPOSITION XVII. THEOREM. If two planes are perpendicular to each other, a line drawn in one of them perpendicular to their common intersection, will be perpendicular to the other plane. D Let the plane AB be perpendicular to NM; then if the line AP be perpendicular to the intersection BC, it will also be perpendicular to the plane NM. For, in the plane MN draw PD perpendicular to PB; then, because the planes are perpendicular, the angle APD is a right angle; therefore, the line AP is perpendicular to the two straight lines PB, PD; therefore it is perpendicular to their plane MN (Prop. IV.). N AR N P E Cor. If the plane AB is perpendicular to the plane MN, and if at a point P of the common intersection we erect a perpendicular to the plane MN, that perpendicular will be in the plane AB: for, if not, then, in the plane AB we might draw AP per M* pendicular to PB the common intersection, and this AP, at the same time, would be perpendicular to the plane MN; therefore at the same point P there would be two perpendiculars to the plane MN, which is impossible (Prop. IV. Cor. 2.). PROPOSITION XVIII. THEOREM. If two planes are perpendicular to a third plane, their common intersection will also be perpendicular to the third plane. The proposition requires demonstration only when the plane angle, which is compared to the sum of the other two, is greater than either of them. Therefore suppose the solid angle S to be formed by three plane angles ASB, ASC, BSC, whereof the angle ASB is A4 the greatest; we are to show that ASB ASC+BSC. AL E If a solid angle is formed by three plane angles, the sum of any two of these angles will be greater than the third. D N B In the plane ASB make the angle BSD=BSC, draw the straight line ADB at pleasure; and having taken SC=SD, draw AC, BC. The two sides BS, SD, are equal to the two BS, SC; the angle BSD=BSC; therefore the triangles BSD, BSC, are equal; therefore BD=BC. But AB AC+BC; taking BD from the one side, and from the other its equal BC, there re mains AD AC. The two sides AS, SD, are equal to the two AS, SC; the third side AD is less than the third se AC; therefore the angle ASD ASC (Book I. Prop. IX. Sch.). Adding BSD-BSC, we shall have ASD+BSD or ASB< ASC+BSC. PROPOSITION XX. THEOREM. The sum of the plane angles which form a solid angle is always less than four right angles. Cut the solid angle S by any plane ABCDE; from O, a point in that plane, draw to the several angles the straight lines AO, OB, OC, OD, OE. S B 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. A formed about the point O. But at the point B the sum of the angles ABO, OBC, equal to ABC, is less than the sum of the angles ABS, SBC (Prop. XIX.); in the same manner at the point C we have BCO+OCD<BCS+SCD; and so with all the angles of the polygon ABCDE: whence it follows, that the sum of all the angles at the bases of the triangles whose vertex is in O, is less than the sum of the angles at the bases of the triangles whose vertex is in S; hence to make up the deficiency, the sum of the angles formed about the point O, is greater than the sum of the angles formed about the point S. But the sum of the angles about the point O is equal to four right angles (Book I. Prop. IV. Sch.); therefore the sum of the plane angles, which form the solid angle S, is less than four right angles. Scholium. This demonstration is founded on the supposition that the solid angle is convex, or that the plane of no one surface produced can ever meet the solid angle; if it were otherwise, the sum of the plane angles would no longer be limited, and might be of any magnitude. PROPOSITION XXI. THEOREM. If two solid angles are contained by three plane angles which are equal to each other, each to each, the planes of the equal angles will be equally inclined to each other. Let the angle ASC=DTF, the angle ASB DTE, and the angle BSC=ETF; then will the inclination of the planes ASC, ASB, be equal to that of the planes DTF, DTE. P Having taken SB at pleasure, draw BO perpendicular to the plane ASC; from the point O, at which the perpendicular meets the plane, draw OA, OC perpendicular to SA, SC; draw AB, BC; next take TE=SB; draw EP perpendicular to the plane DTF; from the point P draw PD, PF, perpendicular respectively to TD, TF; lastly, draw DE, EF. The triangle SAB is right angled at A, and the triangle TDE at D (Prop. VI.): and since the angle ASB=DTE we have SBA=TED. Likewise SB=TE; therefore the triangle SAB is equal to the triangle TDE; therefore SA=TD, and AB=DE. In like manner, it may be shown, that SC=TF, and BC=EF. That granted, the quadrilateral SAOC is equal to the quadrilateral TDPF: for, place the angle ASC upon its equal DTF; because SA=TD, and SC=TF, the point A will fall on D, and the point C on F; and at the same time, AO, which is perpendicular to SA, will fall on PD which is perpendicular to TD, and in like manner OC on PF; wherefore the point O will fall on the point P, and AO will be equal to DP. But the triangles AOB, DPE, are right angled at Ô and P; the hypothenuse AB=DE, and the side AO=DP: hence those triangles are equal (Book I. Prop. XVII.); and consequently, the angle OAB=PDE. The angle OAB is the inclination of the two planes ASB. ASC; and the angle PDE is that of the two planes DTE, DTF; hence those two inclinations are equal to each other. It must, however, be observed, that the angle A of the right angled triangle AOB is properly the inclination of the two planes ASB, ASC, only when the perpendicular BO falls on the same side of SA, with SC; for if it fell on the other side, the angle of the two planes would be obtuse, and the obtuse angle together with the angle A of the triangle OAB would make two right angles. But in the same case, the angle of the two planes TDE, TDF, would also be obtuse, and the obtuse angle together with the angle D of the triangle DPE, would make two right angles; and the angle A being thus always equal to the angle at D, it would follow in the same manner that the inclination of the two planes ASB, ASC, must be equal to that of the two planes TDE, TDF. Scholium. If two solid angles are contained by three plane |