CD: AB = √2: √3 (363); the square constructed upon CD as base is the required square. 641. Let AB be a side of the given pentagon. Find a line CD such that CD: AB=√3: √4; the pentagon constructed upon CD as base is the required pentagon. 642. Let AB be a side of the given hexagon. Find a line CD such that CD: AB =√4: √5; the hexagon constructed upon CD as base is the required hexagon. 643. Let AB be the radius of the given circle. Find a line CD such that CD: AB=√5: √6; the circle described with radius equal to CD is the circle required. NOTE. For solutions to exercises in the Appendix to Plane Geometry bound separately, see p. 105 of this Key. PART II. EXERCISES, pp. 227-257. 644. The revolving plane would coincide with MN on coming to contain the point B. If the plane continued to revolve, it would, after a semi-revolution, again coincide with MN. 645. Such an intersection must be either a broken line or a curve. 'The intersection of a plane with the surface of an apple, for example, is a curve. REMARK. — Illustrate by cutting an apple, or similar object, in two. 646. The converse of Cor. 4, Prop. II., viz., A straight line is the intersection of two planes, is not generally true. Thus the intersection of a plane with any surface on which a straight line can be drawn, such as a cylindrical or conical surface, or the intersection of two such surfaces, may be a straight line. 647. The figure is not a stellate polygon, because the chords do not form a continuous broken line, but two separate triangles. 648. As these planes have the points A and P in common, the line joining these points must lie in each, and be their common intersection. 649. Since AB, AC, AD, are the hypotenuses of right triangles having the arm AP common, the circumferences described must pass through the points A and P, and have AP as a common chord. 650. AD2≈ AP2 + PD2 ≈ AP2 + AB2; AD2 – AP2 = AB2; ··. (AD + AP)(AD – AP)= AB2. .. 651. If planes be passed through DD' and AA', and through DD' and CC', DA', DC will be par'ms; hence DA, DC are || and = to D'A', D'C' respectively; hence D=Z D', and ▲ DAC = ▲ D'A'C'. 652. In the diagram for Prop. XIII., suppose the planes AB', AC', BC", are to MN, the plane of ▲ ABC. Since AA', BB', C'C', are then to MN (475), the & BAC, ABC, ACB, are the plane angles of the dihedral angles whose edges are AA', BB', C'C', respectively; hence no two of these dihedral angles are equal, two are equal, or all are equal according as ▲ ABC is scalene, isosceles, or equilateral. 653. Since FG: AC = DF: DA, and FE:BD = AF: DA, we have FG. DA≈ AC· DF, and FE · DA≈BD · AF; ... FG: FE = AC DF: BD · AF. H 654. Let HH', KK' be the intersections with plane MN of the planes passed through GG'. Since GG' is parallel to BC (Hyp.), GG' is parallel to plane MN (448); .. HH', KK' are parallel to GG' (449); .. HH' is parallel to KK'. 655. In the diagram for Ex. 654, suppose a plane AEPE' perpendicular to MN, cutting GG', BC, HH', KK', in A, P, E, E', respectively. The plane is perpendicular to BC, HH', KK'. Join AE, AE'; then, according M E E E' K C K HN as PE is or is not equal to PE', rt. ▲ APE is or is not equal to rt. ▲ APE', and ▲ PEA is or is not equal to ≤PE'A; i.e., the dihedral angles measured by PEA, PE'A are or are not equal. 656. As shown in the preceding exercise, if BC is equidistant from HH' and KK', rt. ▲ APE = rt. ▲ APE', and Z PAE = L PAE'. 657. KK' being on the same side of BC as HH', let plane APE cut KK' in E', and join AE'; then ZAEE' = ≤ AE'P – ▲ EAE', and these angles are the plane angles that measure the dihedral angles whose edges are HH', KK', and GG', respectively. 658. In the diagram for Prop. XX., if plane PQ is perpendicular to plane RS, BQ is perpendicular to SB, and therefore to plane RS. Similarly BS is perpendicular to plane PQ, while AB is perpendicular to plane MN; hence AB, BQ, BS, are perpendicular to each other. 659. Let the production of QB be BQ'. If ZSBQ = 140°, then ZSBQ'=180°— 140°=40°; ... dihedral ≤ SABQ: dihedral / SABQ' = 140:40 7:2. 660. In the plane of PE and PF, quadrilateral PEOF has & PEO, PFO right angles; points P, E, 0, F, are concyclic (Ex. 398). 661. If FO is equal to FP, PFO is an isosceles right triangle; ... ZFOP = 45°; similarly EOP = 45° ; .·. ▲ EOF = 90°; .. CABD is a right dihedral angle. = : 120°. 662. Suppose FE joined. If FE FP EP, PEF is an equilateral triangle, ▲ EPF = 60°, and ▲ EOF 180° - 60° 663. In the diagram for Prop. XXII., draw AB' parallel to CD. Then CB' is a parallelogram, and AB' = CD. But AB = CD (Hyp.); .. AB=AB', which cannot be the case unless AB coincides with AB and is therefore parallel to CD. Α' C B' B N 664. Let AB, A'B' be parallel lines having projections CD, C'D' upon the same plane MN. (1) If AB, A'B' lie in the same plane to MN, then CD, C'D' coincide. (2) If AB, A'B do not lie in the same plane to MN, CD, C'D' will not coincide, but will be parallel. For otherwise they would meet in some point O, if produced, and from O could be drawn, in their respective planes of projection, lines OE, OE' || to AB, A'B' respectively. OE' being || to A'B' would also be to AB (445), so that through the same point would be drawn two parallels OE, OE', to the same line AB, which is impossible. 665. For the point in which BA produced meets MN, is its own projection, and lies, therefore, in DC produced (481). 666. BA being produced to meet DC produced in E (Ex. 665), we have ED: EC BD: AC = m:n; = ... ED-EC: EC = m - n: n; i.e., CD: EC = m — n: n. 667. (1) All lines in the same plane of projection have their projections in the same straight line, irrespective of what angles they make with their common projection. (2) All lines in parallel planes of projection have their projections parallel (Ex. 664), irrespective of what angles they make with their respective projections. 668. Since the line with which a line oblique to a plane makes the least angle is that part of its projection with which it makes an acute angle (482), the line with which it makes the greatest angle is that part of its projection with which it makes an obtuse angle. For this obtuse angle is the supplement of the least angle the oblique line makes with any line in the plane. 669. Let AB be the line making an angle of 42° with each of the planes PQ, MN, intersecting in CD. From A draw AE to CD, and pass a plane through AB and AE; the plane AEB is to CD and therefore to PQ and MN (470). Hence its intersections with MN and PQ are the projections of AB upon those planes. Now ▲ EAB, having two angles each equal to 42°, has a vertical / AEB equal to 96°, which is the plane angle of dihedral angle CD (462). GEOM. KEY. - 6 D E B 670. There are four trihedral angles formed by PQ and RS with MN, and having the common vertex B. 671. If SBQ is a right angle, then since ABQ, ABS are each right angles (475), the trihedral B-ASQ is trirectangular; and similarly of the other trihedrals. 672. The intersection of a plane with any two edges of the polyhedral angle, forms the base of a triangle and cannot be to both (84). A plane passed to the edge of the dihedral angle formed by any two faces of the polyhedral angle, produced if necessary, will be to each of these faces; and, as seen above, a plane can be to only one edge. 673. Let the lines be A, B, and C. Planes can be passed through, and determined by, A and B, A and C, B and C; i.e., three planes. 674. Let the lines be A, B, C, and D. They can be combined, two and two, in six ways: A and B, A and C, A and D, B and C, B and D, C and D. Hence six planes can be determined by these lines. 675. Let the line AB be parallel to each of the planes MN, PQ, intersecting in CD. Through AB pass a plane to MN and intersecting PQ in EF. Then AB is || to EF (449), and CD is to EF (451); .. AB is || to CD. E D F 676. Let ABC, DEF be two similar triangles, having AB, AC, BC, homologous and || to DE, DF, EF, respectively, but EF < BC. Join AD, BE, CF. Since EF <BC, BE is not || to CF; :. BE, CF will meet, if produced, in a point G. Join GD. Since EF is to BC, GB: GE= BC: EF. But & ABC, DEF being similar, GB: GE=BA: ED; :. GE: ED = GB: BA. Since ED is to BA, ZGED: ZGBA; ..▲ GED is similar to ▲ GBA; :. ≤ GDE = /GAB; :. ZGDE is supplement of ZADE; :. DA is in the same straight line with DG (49). B = 677. For if a plane be passed through the line perpendicular to the first plane, its intersections with all the planes will be parallel, since 679. The points lie in the plane that is perpendicular to the given plane and intersects it in the given line. QUESTIONS, p. 258. 680. One point determines only a point; two, a straight line; three, a plane. 681. To those contained in Props. I. and V. 682. The required locus is the line perpendicular to the plane of the circumference through its center. This may be proved by joining any point in the perpendicular with any two points in the circumference, and these points with the center. 683. The required locus is the line perpendicular to the plane of the triangle through the intersection of the bisectors of any two of its angles. This may be proved by joining any point in the perpendicular with the vertices of those angles; etc. 684. The required locus is the line perpendicular to the plane of the triangle, through the intersection of the perpendiculars drawn to any two of its sides at their respective mid points. This may be proved by joining any point in the perpendicular with those points; etc. 685. To that contained in Prop. XXIII. 686. The required locus is the plane that is perpendicular to the given line at the given point (434). 687. To those contained in Props. XXVI. and XXVII. respectively. 688. The locus is the plane perpendicular to the given plane and intersecting it in the given line. This follows from Arts. 479, 480, 481. 689. The required locus consists of two planes parallel to the given plane, one on each side of it and at the given distance from it. 690. To Axiom 10 and Prop. XXXII. respectively. 691. To that contained in Prop. VIII. 692. To that contained in Prop. V. 693. (1) The locus is the plane passed through the intersection of the given planes and bisecting their dihedral angle (478). (2) The locus is the plane perpendicular to the mid point of any line intercepted between the given planes and perpendicular to them (450). THEOREMS, p. 258. 694. Let the planes AP, BQ intersect in AB; and let CD, which is 1 to AP at D, meet BQ in C. Draw DE Since plane CDE is to plane BQ (470); 1 to AB, and join CE. E B 695. Let ABCD be a quadrilateral in space; i.e., CDB makes with ADB a dihedral angle whose edge is DB. If a plane be passed parallel to ADB and P |