> 130q` ^bjbjqPqP .::/(@@@@@@@@$;h9@@@@@@@?@@@@@41$fV.U0 " $@@@@@@@@@@@@@@dConsider an angle on the sphere made by two tangents to the sphere at a point V. For definiteness let them be labeled VX and VY. Label the center of the sphere C and the South pole S. The triangle SCV is isosceles. The exterior angle at C is and the two equal interior angles are /2. A plane is defined by the two line segments VX and SV; and another by VY and SV. They meet along the line SV and the angle XVY is formed by the intersection of these two planes with the tangent plane of the sphere at the point V. Thus the angle XVY is determined by the angle between the dihedral planes and the angle between SV and the normal to the sphere s tangent plane at V. This latter angle is /2. The projection of the angle XVY on the sphere s equatorial plane is in turn determined by the angle between SV and SC, the normal to the equatorial plane. But this angle is also /2 (actually /2 if measured in a consistent sense in the plane defined by SC and SV), so the projected angle is the same as the original angle XVY.
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A circle on the sphere is defined by the intersection of the sphere with a plane. Define a radius in the direction of the normal to that plane, CV, where V is the point where it meets the sphere. Consider the cone defined by rays from S through the points of the circle. The circle is delineated by the intersection of these rays and the plane of the circle. These rays could also be defined by the figure they form in a plane normal to SV. The circle is then implied by this figure together with the angle between SV and CV, the normal to the plane of the circle. As before, we can call this angle /2, and find that it equals the angle between SV and the normal to the equatorial plane, SC (apart from opposite sign). It follows that the intersection of the cone with the equatorial plane is also a circle.
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