Page "learned" Paragraph 358

To do this we must first show that every line which meets **zg in a point P meets its image at P.

To see this, consider a general pencil of lines containing a general secant of Aj.

By ( 1 ), the image of this pencil is a ruled surface of order Af which is met by the plane of the pencil in a curve, C, of order Af.

On C there is a Af correspondence in which the Af points cut from C by a general line, l, of the pencil correspond to the point of intersection of the image of L and the plane of the pencil.

Since C is rational, this correspondence has K coincidences, each of which implies a line of the pencil which meets its image.

However, since the pencil contains a secant of **zg it actually contains only Af singular lines.

To avoid this contradiction it is necessary that C be composite, with the secant of **zg and a curve of order Af as components.

Thus it follows that the secants of **zg are all invariant.

But if this is the case, then an arbitrary pencil of lines having a point, P, of **zg as vertex is transformed into a ruled surface of order Af having Af generators concurrent at P.

Since a ruled surface of order N with N concurrent generators is necessarily a cone, it follows finally that every line through a point, P, of **zg meets its image at P, as asserted.