Nie. As in Euclid. But we prove what Euclid has assumed as an Axiom, namely, that two right Lines cannot have a common segment.
Min. I am glad to hear you assert that Euclid has assumed it 'as an Axiom,' for the interpolated and illogical corollary to Euc. I. 11 has caused many to overlook the fact that he has assumed it as early as Prop. 4, if not in Prop. 1. What is your proof?
Niemand reads.
'Two straight Lines cannot have a common segment.'
![](http://upload.wikimedia.org/wikipedia/commons/thumb/2/2f/Euclid_and_His_Modern_Rivals_Page065.png/250px-Euclid_and_His_Modern_Rivals_Page065.png)
'For if two straight Lines ABC, ABH could have a common segment AB; then the straight Line ABC might be turned about its extremity A, towards the side on which BH is, so as to cut BH; and thus two straight Lines would enclose a space, which is impossible.' Min. You assume that, before C crosses BH, the portions coinciding along AB will diverge. But, if ABH is a right Line, this will not happen till C has passed H.
Nie. But you would then have one portion of the revolving Line in motion, and another portion at rest.
Min. Well, why not ?
Nie. We may assume that to be impossible ; and that,
F