If BPC, DPE are circular sections of the cone vertex A, a sphere can be drawn containing these two sections, and the cone will touch the sphere at the points P, P
1 where the line of intersection of the two planes meets the sphere. Thus AP
2 = AD AB. Taking BPC as the base of the cone, if AC is the least and AB the greatest of the generating lines drawn from A to the base, ABC will be a principal section of the cone and the planes of circular ection of the cone are perpendicular to the plane ABC. If AF is perpendicular to DE and AH perpendicular to BC,
2 = AF AH sin B sin C;
That is, AP
2 is the product of the perpendiculars drawn from P to the planes hrough the vertex A parallel to the planes of circular section, divided by in B sin C.