Elements of Geometry

cutting CDB in EF, EF will be parallel to DB (451); hence CE : ED = CF : FB.

966. Let ABCD, as in Ex. 695, be a quadrilateral in space, and E, F, G, H, the mid points of CD, CB, AD, AB, respectively. In the plane passed through EF, GH, suppose BEG, FH joined. Then since EF, GH are parallel to DB and equal to DB, EF is parallel and equal to GH; .:. EGFH is a parallelogram (142).

697. Let AB, CD, EF, be any three of the parallel intersections, and P be the given point. Through P draw QR || to those intersections, and through Ppass a plane I to QR. This plane will be I to AB, CD, EF, etc., since these lines are I to it (444), and therefore to the planes of which these are the intersections (470); hence all the perpendiculars from P to these planes will lie in that plane.

698. In the diagram for Ex. 694, suppose CD equally inclined to AP and BQ. Through CD pass a plane I to AB and cutting it in E. This plane, being I to both AP and BQ (470), contains the projections CE, DE of CD upon BQ, AP resp. Now ∠ECD = ∠ EDC (Hyp.); .·. ED = EC. 699. A being without the plane, let AP make equal angles with the lines PB, PC, PD, in plane MN. With Pas center, describe a circumference cutting PB, PC, PD, in E, F, G, respectively, and join AE, AF, AG. Since the angles Nat Pare equal (Hyp.), AE=AF=AG

(66, 70). Now the locus of all points at a distance AE from A is a circumference whose center is the foot of the perpendicular from A, and P is the center of the only circumference that can pass through E, F, and G (185); hence AP is to MN. 700. Let the plane MN intersect the parallelogram ABCD in the diagonal BD, and let AA', CC' be perpendiculars from A and C to plane MN. Join A' and C with O, the intersection of diagonal AC with BD. AA' is || to CC' (443); .. ZOAA'=∠OCC' (112); also hypot. OA= hypot. OA'; .·. AA' = CC' (73, 70).

701. Let AP, BQ be planes intersecting in AB, and from O, a point between them, draw OD, OC I to AP, BQ respectively. Through OC, OD, pass a plane intersecting AP, BQ in DE, CE respectively. This plane is I to AB (476); ... DEC is the plane angle of dihedral QABP, and is the supplement of 20, since in quadrilateral OCED, A D and C are right angles. Hence ∠O is the equal (44) of the plane angle of the dihedral angle formed by producing AP through AB.

702. For the plane I to the common intersection at any point will contain all the I's to the intersection at that point (434).

703. Let SA, SB, SC, be the edges of trihedral LS-ABC. Through SA, SB pass planes bisecting the dihedral A SA, SB respectively, and let these planes intersect in the straight line SD. Take any point Pin SD. Since Pis in the bisecting plane of dihedral ZSA, Pis equidistant from 4

the faces ASB, ASC. For a similar reason Pis equidistant from the faces BSC and ASB. Hence, being equidistant from ASC and BSC, P must lie in the bisecting plane of dihedral SC; that is, any point in SD lies in the three bisecting planes.

704. On the edges of the trihedral ∠S-ABC lay off SA, SB, SC, and through A, B, C, pass a plane forming, by its intersection with the faces, the ABC. Draw AA' I to BC, and pass a plane through SA, AA'; this plane is to SBC (470). Similarly planes A passed through SB, BB' and SC, CC are 1 to SAC, SAB respectively, BB', C'C' being to AC, AB respectively. But AA', BB', СС", intersect in a point O (Ex. 152); ... S and O being common to the three planes, the line joining So is their common intersection.

705. In the diagram for Ex. 704, suppose SA = SB = SC; then the ▲ SAB, SAC, SBC, are isosceles, and the bisectors of the face angles are also the medians of those triangles. Hence the planes passed through SA, SB, SC, and the bisectors of the opposite face angles, intersect ABC in the medians to BC, AC, AB. But these medians intersect in a point O (Ex. 154); .. S and O being common to the three planes, the line joining SO is their common intersection.

706. In the diagram for Ex. 705, since ASAB, SAC, SBC, are isosceles, the bisectors of the face angles are I to BC, AC, AB, at their respective mid points. Hence the planes passed through the bisectors to the faces are also I to BC, AC, AB, and their inter

sections with △ ABC form the perpendiculars at the mid points of its sides. But these perpendiculars meet in a point O (155); ... Sand O being common to the three planes, SO is their common intersection.

LOCI, p. 259.

707. The plane passed perpendicular to the line joining the given points A, B, at the mid point of AB, is the required locus; for the line joining that mid point C with any other point in the plane is perpendicular to AB,; hence that point is equidistant from A and B (96).

708. The two planes passed perpendicular to the plane of the given lines, through the bisectors of the angles formed by those lines, constitute the required locus. For the foot of the perpendicular from any point in either of those planes to the bisector through which it passes, is equidistant from the given lines (101); hence that point

is also equidistant from the given lines.

709. (1) Let the planes intersect. Pass a plane BAC perpendicular to the intersection of the given planes; then find (Ex. 457) the lines, such as AD, that constitute the locus of a point whose distances from AB, AC are in the given ratio. The planes passed through the intersection of the given planes and AD constitute the required locus. For the distances of any point D in either of those planes from the given planes are in the given ratio. (2) If the lines are parallel, we proceed similarly by means of Ex. 457.

710. Draw perpendiculars to any two sides of the triangle at their mid points, and let them intersect in O. The perpendicular to the plane of the triangle through O is the required locus; for O being equidistant from the vertices of the triangle, any point in the perpendicular to the plane of the triangle through O is equidistant from the vertices. 711. The perpendicular to the plane of the given triangle through the intersection of any two of the angles is the locus required (154). 712. The perpendicular to the plane of the quadrilateral through the center of the circumference that can be described about it (Ex. 398), is the required locus. For that center is equidistant from the vertices, which are concyclic.

713. The perpendicular to the plane of the circle through its center is the required locus.

714. If the given planes are parallel, it is evident that no point can be equidistant from all three. But if one plane intersect the other two, the required locus will consist of the intersections of the planes that bisect the dihedral angles formed by the one plane with the other two. For any point O in either intersection will be equidistant from the three given planes (478).

715. Let S-ABC be the trihedral angle. The required locus is the intersection of the planes passed perpendicular to any two of the faces through the bisectors of their respective face angles. For any point in that intersection is equidistant from all the edges (Ex. 706).

716. (1) If the given planes intersect, the planes bisecting their dihedral angles constitute the locus of points equidistant from those planes. The plane perpendicular at the mid point of the line joining the given points, is the locus of points equidistant from the given points. The line or lines formed by the intersection of these two loci constitute the required locus. (2) If the given planes are parallel, the required locus will consist of the intersection of the plane equidistant from the given planes with the plane that is the locus of points equidistant from the given points.

PROBLEMS, p. 260.

717. Through any point Cin the given line AB lying in plane MN, pass a plane PQ perpendicular to AB, and intersecting MN in CD. In PQ draw a line CE, making with CD an ∠ECD equal to the given angle. The plane passed through AB, CE is the plane required.

718. Through any point Cin the given line AB pass a plane PQI to AB, and intersecting the given plane MN in DE. In PQ drawa. line CF, making with DE an ∠CFE equal to the given angle; etc.

719. Pass a plane R perpendicular to the given edge at any point A; R will be perpendicular also to the faces P, Q of the given dihedral angle (470). Bisect the plane angle formed by the intersection of R with Pand Q. The plane passed through the bisector and the given edge will be the plane required.

720. From the given point A draw AB I to P, the given plane, and through A pass a plane I to AB. This plane Qis || to plane P.

721. At any point A in the given plane P, draw AB perpendicular to P. On AB lay off AC equal to the given distance, and through C pass a plane perpendicular to AB; etc.

722. Through AB, the given line, pass two planes, one of which passes through C, the given point. In the plane containing C, draw CD perpendicular to AB, and in the other plane draw DE perpendicular to AB. The plane of CD and DE is the plane required.

723. Let S-ABC be the trihedral angle. Through the bisectors of two of the face angles, pass planes I to those faces; the plane passed

through the vertex SI to the intersection SO is the plane required. For through any point O pass a plane I to SO, so as to intersect the faces in AB, AC, BC. Join OA, OB, OC. The rt. A SOA, SOB,

C
E

SOC, are equal; ... A ASO, BSO, CSO, are equal; ... the made by SA, SB, SC, with the I plane through S are equal (44). 724. (1) Let the given points A, B be on the same side of the given plane MN. Through the points A and B pass a plane PI to MN, and intersecting it in EE'. In plane P draw AC to EE', and produce to D, so that CD = AC. Draw BD, cutting EE' in X. X is the required point (Ex. 248). (2) If A and B are on opposite sides of the plane, we proceed by a construction similar to that given above (Ex. 247).

725. Let A, B, C, be the given points, and MN the given plane. Find the locus of points equidistant from A, B, and C (Ex. 710); the point X, in which this locus meets MN, is the required point. If the plane of ABC is perpendicular to the given plane, there can be no point such as is required.

726. Find the locus of points equidistant from the given points. The point in which this plane cuts the given line is the required point. If the plane is || to the given line, there can be no point such as is required.

EXERCISES, pp. 263-280.

727. The sum of the lateral dihedral angles = (5 – 2)180° = 540°. 728. The prism MN has 5+2 = 7 faces; 4×5+2× 5 = 30 face angles; 5+5 × 2 = 15 dihedral angles; 5 × 2 = 10 trihedral angles. 729. If the base of MN has n sides, the prism has (n+2) faces; (4n+2n)=6n face ; (n + 2n)=3n dihedral 4, and 2 n trihedral 4. 730. Since AC'≈ Ac' (525), AC'− (ABCd – C')≈ Ac' - (ABCd – C'); i.e., AA'D'D – d' ≈ BB'C'C – c'.

731. The solid angle whose vertex is A is a tetrahedral angle, having as faces A'Aa, A'AD, DAd, and dAa. The angle whose vertex is Dis trihedral.

732. Let abc, dbc be respectively an isosceles right triangle, and an equilateral triangle, having the hypotenuse be of abc as common base. Join da, and produce to meet bc in e; de is perpendicular to be at its mid point (74). The triangles, having the same base, are as their altitudes. Now ec : de = 1 : √3, and ec : ae = 1:1;.. de:ae = √3:1. The prisms, having equal altitudes, are as their bases; ... prism A: prism B = 1: √3.

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Elements of Geometry: Plane and Solid