§ 95. It remains to be proved that every point S on the surface may be taken as centre of one of the pencils which generate the surface. Let S be any point on the surface Φ′ generated by the reciprocal pencils S1 and S2. We have to establish a reciprocal correspondence between the pencils S and S1, so that the surface generated by them is identical with Φ. To do this we draw two planes α1 and β1 through S1, cutting the surface Φ in two conics which we also denote by α1 and β1. These conics meet at S1, and at some other point T where the line of intersection of α1 and β1 cuts the surface.
In the pencil S we draw some plane σ which passes through T, but not through S1 or S2. It will cut the two conics first at T, and therefore each at some other point which we call A and B respectively. These we join to S by lines a and b, and now establish the required correspondence between the pencils S1 and S as follows:—To S1T shall correspond the plane σ, to the plane α1 the line a, and to β1 the line b, hence to the flat pencil in α1 the axial pencil a. These pencils are made projective by aid of the conic in α1.
In the same manner the flat pencil in β1 is made projective to the axial pencil b by aid of the conic in β1, corresponding elements being those which meet on the conic. This determines the correspondence, for we know for more than four rays in S1 the corresponding planes in S. The two pencils S and S1 thus made reciprocal generate a quadric surface Φ′, which passes through the point S and through the two conics α1 and β1.
The two surfaces Φ and Φ′ have therefore the points S and S1 and the conics α1 and β1 in common. To show that they are identical, we draw a plane through S and S2, cutting each of the conics α1 and β1 in two points, which will always be possible. This plane cuts Φ and Φ′ in two conics which have the point S and the points where it cuts α1 and β1 in common, that is five points in all. The conics therefore coincide.
This proves that all those points P on Φ′ lie on Φ which have the property that the plane SS2P cuts the conics α1, β1 in two points each. If the plane SS2P has not this property, then we draw a plane SS1P. This cuts each surface in a conic, and these conics have in common the points S, S1, one point on each of the conics α1, β1, and one point on one of the conics through S and S2 which lie on both surfaces, hence five points. They are therefore coincident, and our theorem is proved.
§ 96. The following propositions follow:—
A quadric surface has at every point a tangent plane.
Every plane section of a quadric surface is a conic or a line-pair.
Every line which has three points in common with a quadric surface lies on the surface.
Every conic which has five points in common with a quadric surface lies on the surface.