Let MN, PQ, be two parallel planes, and let AB be perpendicular to NM; then will it also be perpendicular to QP. Having drawn any line BC in the plane PQ, through the lines AB and BC, draw a plane ABC, intersecting the plane MN in AD; the intersection AD will be parallel to BC (Prop. X.); but the line AB, being perpendicular to the plane MN, is perpendicular to the straight line AD; therefore also, to its parallel BC (Book I. Prop. XX. Cor. 1.): hence the line AB being perpendicular to any line BC, drawn through its foot in the plane PQ, is consequently perpendicular to that plane (Def. 1.). Cor. Hence it follows, that two parallel planes are every where equidistant: for, suppose EG were perpendicular to the plane PQ; the parallel FH would also be perpendicular to it (Prop. VII.), and the two parallels would likewise be perpendicular to the plane MN (Prop. XI.); and being parallel, they will be equal, as shown by the Proposition. PROPOSITION XIII. THEOREM. If two angles, not situated in the same plane, have their sides parallel and lying in the same direction, those angles will be equal and their planes will be parallel. Let the angles be CAE and DBF. Make AC=BD, AE= BF; and draw CE, DF, AB, CD, EF. Since AC is equal and parallel to BD, the figure ABDC is a parallelogram; therefore CD is equal and parallel to AB. For a similar reason, EF is equal and parallel to AB; hence also CD is equal and parallel to EF; hence the figure CEFD is a parallelogram, and the side CE is equal M P A and parallel to DF; therefore the triangles CAE, DBF, have their corresponding sides equal; therefore the angle CAE= DBF. Again, the plane ACE is parallel to the plane BDF. For suppose the plane drawn through the point A, parallel to BDF, were to meet the lines CD, EF, in points different from Cand E, for instance in G and H; then, the three lines AB, GD, FH, would be equal (Prop. XII.): but the lines AB, CD, EF, are already known to be equal; hence CD-GD, and FH=EF, which is absurd; hence the plane ACE is parallel to BDF. Cor. If two parallel planes MN, PQ are met by two other planes CABD, EABF, the angles CAE, DBF, formed by the intersections of the parallel planes will be equal; for, the inter section AC is parallel to BD, and AE to BF (Prop. X.); therefore the angle CAE=DBF. PROPOSITION XIV. THEOREM. If three straight lines, not situated in the same plane, are equal and parallel, the opposite triangles formed by joining the extremities of these lines will be equal, and their planes will be parallel, If two straight lines be cut by three parallel planes, they will be divided proportionally. 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: we are now to show that 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 M P B E AE EB AG: GD; N in like manner, the intersections AC, GF, being parallel, AG: GD :: CF : FD; the ratio AG: GD is the same in both; hence AE EB:: CF: FD. PROPOSITION XVI. THEOREM. If a line is perpendicular to a plane, every plane passed through the perpendicular, will also be perpendicular to the plane. 21 B D 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 perpendicu lars 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. 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. 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 perpendicu lar, the angle APD is a right angle; therefore, the line AP is perpendicular to the two straight M E B D N lines PB, PD; therefore it is perpendicular to their plane MN (Prop. IV.), 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 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. If a solid angle is formed by three plane angles, the sum of any two of these angles will be greater than the third. 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 A the greatest; we are to show that ASB ASC+BSC. 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 BDBC. But AB<AC+BC; taking BD from the one side, and from the other its equal BC, there re |