Page:Carroll - Euclid and His Modern Rivals.djvu/292

254 . II. 6.

A Pair of separational Lines are equidistantial from each other.



Let AB, CD be separational Lines : it shall be proved that they are equidistantial from each other.

In AB take any 2 points E, F; and draw EG, FH, ⊥ CD.

Now, if FH > EG, from it cut off KH equal to EG; and join EK;

then, ∵ EG = KH,

∴ EK, CD have a common perpendicular;

∴ EK is separational from CD;

∴ AB, EK, intersecting Lines, are both separational from CD ; which is absurd ;

∴ FH is not > EG.

Similarly it may be proved that EG is not > FH.

Therefore EG = FH.

Similarly it may be proved that any 2 points in AB are equidistant from CD, and that any 2 points in CD are equidistant from AB.

Therefore AB, CD are equidistantial from each other.

Therefore a Pair &c.

. II. 11.

A Pair of Lines, which are equally inclined to a certain transversal, are equidistantial from each other.

A Pair of Lines, which are equally inclined to a certain transversal, are separational;

also a Pair of separational Lines are equidistantial from each other;

∴ a Pair of Lines, &c.