Nie. We do not: but I fear that will not mend matters, as we assume, in the course of this Theorem, that a Line can be drawn through a given point, so as to have the same direction as a given Line.
Min. Then we need not examine it further: it must perish with the faulty Axiom on which it rests. What is your next Theorem?
Nie. It answers to Euc. I, 16, 17, and is proved by the Theorem you have just rejected.
Min. Then I must reject its proof, but I will grant you the Theorem itself, if you like, as we know it can be proved from undisputed Axioms. What comes next?
Niemand reads.
P. 14. Th. 13. 'If a Transversal meet two Directives, and make angles with them, the External greater than the Internal, or the sum of the two Internal angles less than two right angles, the two directives must meet.'
Min. A proof for Euclid's Axiom? That is interesting.
Niemand reads.
'For, suppose they do not meet. Then, they should be sepcodal'
Min. (interrupting) 'Should be sepcodal'? Does that mean that they are sepcodal?
Nie. Yes, I think so.
Min. That is, you assume that separational Lines have the same direction?
Nie. We do.
Min. A fearful assumption! (A long silence) Well?
