Consider a circle of radius R in the xy-plane. Its equation is

x² + y² = R²

Now choose a point P in the interior of the circle with coordinates (x_p, y_p). The equation for a line B through P with a slope of b is given by

(y−y_p)/(x−x_p) = b
and hence
y = y_p + b(x−x_p)

The proposition is that the angles between the line and the normals to the circle at the points of intersection are equal.

The proposition can be proven by a brute force computation. The points of intersection can be found by replacing y in the equation for the circle with y_p + b(x−x_p) and solving the resulting quadratic equation for x, the x-coordinate of the point of intersection. There generally will be two solutions. The normals to the circle at any point are equivalent to the radial vectors from the center point to the points of intersection. There however is a more elegant proof.

For the line B construct a line N through the circle center that is perpendicular to B. Let Q=(x_q, y_q) be the point of intersection of the perpendicular with the line B. The line N, because it passes through the center divides the circle into two identical semicircles. In particular the angles labled θ formed by the center, the point Q and the points of intersection of the line B with the circle are in fact equal.

This implies that the other angles, labled φ in the diagram below, are also equal.

The Case of a Sphere (a 3-Sphere)

Consider a sphere of radius R centered at the origin of the coordinate system. The Line B and the center of the sphere determine a plane, call it N. If the line B passes through the origin the plane containing the center and the line is not uniquely determined, but for this case the angles between the line and the radial lines at the points of intersection are both zero so the proposition holds.

From hereon it is assumed that the line does not pass through the center and thus the plane containing the center and the line is uniquely determined. This plane intersects the sphere in a circle of radius R. The radial lines from the origin to the points of intersection are in the plane N. in the plane N. The line B is in the plane N. The angles in question are the same as those for the circle. Thus by the previous case the angles are equal.

The General Case of the n-Sphere

The coordinates for the n-Sphere of radius R are the sets of n numbers {x₁, x₂, …, x_n} such that

x₁² + x₂² + … + x_n² = R²

As in the previous cases, the center of the n-sphere and the line B determine a plane. The coordinate system can be rotated such that the plane corresponds to the plane x_n=0. The intersection of this plane with the n-sphere is simply the (n-1)-sphere

x₁² + x₂² + … + x_n-1² = R²

The line B lies in the plane and intersects the (n-1)-sphere. The center of the (n-1)-sphere is the origin of the coordinate system. This center and the line B determine a plane. The coordinate can be rotated to make the plane coincide with the plane x_n-1=0. This process can be continued until the intersection of the plane determined by the origin and the line B is a 2-sphere. From the case for the 2-sphere the proposition concerning the angles between the line B and the radial lines from the origin holds.