(γ). 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; [Lemma 1.
∴ EK is separational from CD; [Euc. I. 27.
∴ AB, EK, intersecting Lines, are both separational from CD ; which is absurd ; [(β)
∴ 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. Q. E. D.
(δ). 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; [Euc. I. 27.
also a Pair of separational Lines are equidistantial from each other; [(γ).
∴ a Pair of Lines, &c. Q. E. D.
