PROPOSITION V. THEOREM. 697. A sphere may be inscribed in any given tetrahedron. Let A B C D be the given tetrahedron. Bisect the dihedral is at the edges A B, BC and A C by the planes 0 A B, O B C and O A C respectively. Every point in the plane 0 A B is equally distant from the faces A B C and A B D, . § 477 For a like reason, every point in the plane 0 B C is equally distant from the faces A B C and D BC; and every point in the plane 0 A C is equally distant from the faces A B C and ADC. .O, the common intersection of these three planes, is equally distant from the four faces of the tetrahedron. is a sphere described with O as a centre, and with the radius equal to the distance of 0 to any face, will be tangent to each face, and will be inscribed in the tetrahedron. § 673 Q. E. D. 698. COROLLARY. The six planes which bisect the six dihedral angles of a tetrahedron intersect in the same point. ON DISTANCES MEASURED ON THE SURFACE OF THE SPHERE. 699. DEF. The distance between two points on the surface of a sphere is understood to be the arc of a great circle joining the points, unless otherwise stated. 700. DEF. The distance from the pole of a circle to any point in the circumference of the circle is the Polar distance of the circle. PROPOSITION VI. THEOREM. 701. The distances measured on the surface of a sphere from all points in the circumference of a circle to its pole are . P We are to prove arcs P A, PB, PC equal. The rectilinear lines PA, P B and PC are equal, $ 450 (oblique lines drawn from a point to a plane at equal distances from the foot of the I are equal) ; (in equal o equal chords subtend equal arcs). Q. E. D. 702. COROLLARY 1. The polar distance of a great circle is a quadrant. Thus, arcs P A', PB', P'A', P'B', polar distances of the great circle A' B' C D', are quadrants ; for they are the measures of the right angles A' OP, B'OP, A' O P', BO P', whose vertices are at the centres of the great circles P A'P'C', PB' P'D'. 703. Cor. 2. If a point P on the surface of the sphere be at the distance of a quadrant from the two points d' and B' of an arc of a great circle, it is the pole of that arc. For, since the arcs P A and P B' are quadrants, the angles PO A' and P O B' are right angles. And hence, the radius 0 P is perpendicular to each of the lines 0 A' and 0 B', and therefore perpendicular to the plane of the arc A' B' ($ 449). Therefore P is the pole of arc A' B'. § 683 PROPOSITION VII. PROBLEM. 704. To pass a circumference of a great circle through any two points on the surface of a sphere. h Pia Let A and B be any two points on the surface of a sphere. It is required to pass a circumference of a great circle through A and B. From A as a pole, with an arc equal to a quadrant, strike an arc ab, and from B as a pole, with the same radius, describe an arc cd, intersecting a b at P. Then a circumference described with a quadrant arc, with P as a pole, will pass through A and B and be the circumference of a great circle. Q. E. F. 705. COROLLARY. Through any two points on the surface of a sphere, not at the extremities of the same diameter, only one circumference of a great circle can be made to pass. 706. SCHOLIUM. By means of poles arcs of circles may be drawn on the surface of a sphere with the same facility as upon a plane surface, and, in general, the methods of construction in Spherical Geometry are similar to those of Plane Geometry. Thus we may draw an arc perpendicular to a given spherical arc, bisect a given spherical angle or arc, make a spherical angle equal to a given spherical angle, etc., in the same way that we make analogous constructions in Plane Geometry. PROPOSITION VIII. THEOREM. 707. The shortest distance on the surface of a sphere between any two points on that surface is the arc, not greater than a semi-circumference, of the great circle which joins then. Let A B be the arc of a great circle which joins any two points A and B on the surface of a sphere; Let P be any point in AC PQ B. Pass arcs of great circles through A and P, and P and B. $ 704 Join A, P and B with the centre of the sphere 0. The SA OB, A O P and PO B are the face of the trihedral Z whose vertex is at 0. The arcs A B, A P and P B are measures of these . § 202 Now ZAOB< ZAOP+ 2 POB, § 487 (the sum of any two fuce is of a trihedral is > the third Z). i. arc A B < arc A P + arc P B. In like manner, joining any point in AC P with A and P by arcs of great , their sum would be greater than arc AP; and, joining any point in P Q B with P and B by arcs of great ©, the sum of these arcs would be greater than arc P B. If this process be indefinitely repeated the distance from A to B on the arcs of the great © will continually increase and approach to the line A C P Q B. .. arc A B < ACPQ B. Q. E. D. PROPOSITION IX. THEOREM. 708. Every point in an arc of a great circle which bisects a given arc at right angles is equally distant from the extremities of the given arc. Let arc C D bisect arc A B at right angles. We are to prove any point O in CD is equally distant from A and B. Since great circle C D E bisects arc A B at right angles, it also bisects chord A B at right angles. Hence, chord A B is I to the plane C D E at K. .O K is I to chord A B at its middle point. § 430 .. rectilinear lines 0 A and 0 B are equal. $ 58 i. arcs 0 A and 0 B are equal. $ 182 Q. E. D. ChordB at Seat circle from PROPOSITION X. PROBLEM. 709. To pass the circumference of a small circle through any three points on the surface of a sphere. Let A, B and C be any three points on the surface of a sphere. It is required to pass the circumference of a small circle through the points A, B and C. Pass arcs of great circles through A and B, A and C, B and C. § 704 Arcs of great circles ao and bo I to A C and B C at their middle points intersect at o. Then o is equally distant from A, B and C. § 708 .. the circumference of a small circle drawn from o as a pole, with an arc o A will pass through A, B and C, and be the circumference required. Q. E. D. |