Non-Euclidean Geometry

Cover Non-Euclidean Geometry
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It will lie in the projecting plane OFF'. AB, being perpendicular to FF' and to FK, is perpen- BOUNDARY-SUEFACES 51 dicular to this plane, OFF', and therefore to OF. In the same way we prove that AC is perpendicular to OE. Therefore, ^C is perpendicular to OD (Chap. I, I, 5). But OD is the intersection of the plane ABC with the plane ODD'. Hence, BC is perpendicular to this plane and to DD' (Chap. I, II, 15).
DD' being parallel to BB' lies -in the plane determined by BB' and BC, and in this pla
...ne only one perpendicular can be drawn to BC at its middle point. Therefore, if we pass any plane through BB' and from B draw a chord to any other point, C, of its intersection with the surface, the perpendicular in this plane to BC, erected at the middle point of BC, will be parallel to BB'. This proves that the section is a boundary- curve, having BB' for axis, and that the surface can be gener- ated by the revolution of such a boundary -curve around BB'.
Therefore, BB' may be regarded as axis of the surface.


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