2. If the circle is a great circle, what part of a circumference is its polar distance?

3. By passing planes form two equal circles of the same or of equal spheres. How do their polar distances compare ?

4. Select a point on the surface of a sphere which is at a quadrant's distance from each of two other points. Where is this point situated with reference to a pole of a great circle that passes through the other two points?

Theorem. All points in the circumference of a circle of a sphere are equally distant from a pole of the circle.

Data: Any circle of a sphere, as ABC, and its poles, P and P'. To prove all points in the circumference of ABC equally distant from P and also from P'.

Proof. Draw great circle arcs from P to any points in the circumference of ABC, as A, B, and C.

§§ 649, 652,

.. § 450,

hence, § 196,

PP' ABC at its center; chords PA, PB, and PC are equal;

arcs PA, PB, and PC are equal.

In like manner, arcs P'A, P'B, and P'C may be proved equal. But A, B, and C are any points in the circumference of ABC. Hence, § 661, all points in the circumference of ABC are equally distant from P and also from P'.

Therefore, etc.

Q.E.D.

664. Cor. I. The polar distance of a great circle is a quadrant.* 665. Cor. II. The polar distances of equal circles on the same, or on equal spheres, are equal.

*In Spherical Geometry the term quadrant generally means the quadrant of a great circle.

666. Cor. III. A point, which is at the distance of a quadrant from each of two other points on the surface of a sphere, is a pole of the great circle passing through those points.

667. Sch. I. By using the facts demonstrated in § 663 and in § 664 we may draw the circumferences of

small and great circles on the surface of a material sphere.

To draw the circumference of a circle, take a cord equal to its polar distance, and, placing one end of it at the pole, cause a pencil held at the other to trace the circumference, as in the figure.

To describe the circumference of a great circle, a quadrant must be used as the arc.

668. Sch. II. By means of § 666 we are enabled to pass the circumference of a great circle through any

two points, as A and B, on the surface of a material sphere in the following manner:

From each of the given points, as poles, and with a quadrant arc, draw arcs to intersect, as at 0. The circumference described from this intersection with a quadrant arc will be the circumference required.

669. 1. If a plane is perpendicular to a radius of a sphere at its extremity, how many points do the sphere and the plane have in common? What name is given to such a plane?

2. How many points do a straight line and a sphere have in common, if the line is perpendicular to a radius of the sphere at its extremity? What name is given to such a line?

3. What is the direction of every plane or line, that is tangent to a sphere, with reference to the radius drawn to the point of contact?

4. If a straight line is tangent to any circle of a sphere, how does it lie with reference to a plane tangent to the sphere at the point of contact? 5. If a plane is tangent to a sphere, what is the relation to the sphere of any line drawn in that plane and through the point of contact?

6. If two straight lines are tangent to a sphere at the same point, what is the relation of the plane of those lines to the sphere?

Theorem. A plane perpendicular to a radius of a sphere at its extremity is tangent to the sphere.

Data: Any sphere, and a plane, as MN, perpendicular to a radius, as OP, at its extremity P.

To prove

MN tangent to the sphere.

Proof. Take any point except P in MN, as 4, and draw OA.

point 4 is without the sphere.

But 4 is any point in MN except P;

every point in MN except P is without the sphere.

Hence, § 638,

MN is tangent to the sphere at P.

Q.E.D.

670. Cor. I. Any straight line perpendicular to a radius of a sphere at its extremity is tangent to the sphere.

671. Cor. II. Any plane or line tangent to a sphere is perpendicular to the radius drawn to the point of contact.

672. Cor. III. A straight line tangent to any circle of a sphere lies in the plane tangent to the sphere at the point of contact.

673. Cor. IV. Any straight line drawn in a tangent plane and through the point of contact is tangent to the sphere at that point. 674. Cor. V. Any two straight lines tangent to a sphere at the same point determine the tangent plane at that point.

Proposition V

675. Select any four points not in the same plane; form a tetrahedron of which these points are the vertices; and at the centers of the circles circumscribed about any two of its faces erect perpendiculars. How do the distances from any point in either perpendicular to the vertices of the face to which it is perpendicular compare? If these per

pendiculars intersect, how, then, do the distances from their intersection to the four given points compare? Is there any other point that is equidistant from the four given points? What surface, then, may be passed through these four points? How many such surfaces may be passed through them?

Theorem. Through any four points not in the same plane one spherical surface may be passed, and only one. Data: Any four points not in the same

plane, as A, B, C, D.

To prove that one spherical surface may be passed through A, B, C, D, and only one.

Proof.

Suppose H and G to be the centers of circles circumscribed about the triangles BCD and ACD, respectively.

Draw HK plane BCD, and GE plane ACD.

E

B

H

K

A

§ 450, every point in HK is equidistant from points B, C, D, and every point in GE is equidistant from points A, C, D.

From H and G draw lines to L, the middle point of CD.

.. § 444, the plane through HL and GL is perpendicular to CD, and, § 483, this plane is perpendicular to planes BCD and ACD. Const.,

.. § 481,

GE

plane ACD at G;

GE lies in the plane HLG.

In like manner it may be shown that HK lies in plane HLG. Hence, the perpendiculars HK and GE lie in the same plane, and, being perpendicular to planes which are not parallel, they must intersect at some point, as at 0.

Since is in the perpendiculars HK and GE, it is equidistant from B, C, D, and from A, C, D.

Hence, o is equidistant from A, B, C, D, and the surface of the sphere, whose center is 0 and radius OB, will pass through the points A, B, C, D.

Now, § 450, the center of any sphere whose surface passes through the four points A, B, C, D must be in the perpendiculars HK and GE.

Hence, 0, the intersection of HK and GE, must be the center of the only sphere whose surface can pass through A, B, C, D. Therefore, etc.

Q.E.D.

676. Cor. I. A sphere may be circumscribed about any tetrahedron.

677. Cor. II. The four perpendiculars to the faces of a tetrahedron through their centers meet at the same point.

Proposition VI

€78. Form any tetrahedron and pass planes bisecting any three of its dihedral angles which have one face in common. How do the distances from the point of intersection of these bisecting planes to the faces of the tetrahedron compare? What figure, then, may be inscribed in any tetrahedron ?

Theorem. A sphere may be inscribed in any tetrahedron.

To prove that a sphere may be inscribed in D-ABC.

Proof. Bisect any three of the dihedral angles which have one face common, as AB, BC, AC, by the planes OAB, OBC, OAC, respectively.

By § 488, every point in the plane OAB is equidistant from the faces ABC and ABD.

Also every point in plane OBC is equidistant from the faces ABC and BCD; and every point in the plane OAC is equidistant from the faces ABC and ACD.

Therefore, point 0, the intersection of these three planes, is equidistant from the four faces of the tetrahedron.