Now EG = FH; [hyp.
hence, if the diagram be reversed, and so placed on its former traces that G coincides with H, and H with G, K retaining its position, GE coincides with HF, and HF with GE;
∴ E coincides with F, and F with E;
∴ L retains its position;
∴ ∠GKL coincides with ∠HKL, and is equal to it;
∴ ∠s at K are right.
Similarly ∠s at L are right.
Therefore a Pair of Lines, &c. Q. E. D.
(α). II. 1.
A Pair of separational Lines are equally inclined to any transversal.
[N.B. The Contranominal of this will be proved at the end of the series.]
(β). II. 16 (a).
Two intersecting Lines cannot both be separational from the same Line.
![](http://upload.wikimedia.org/wikipedia/commons/thumb/e/e1/Euclid_and_His_Modern_Rivals_Page253.png/220px-Euclid_and_His_Modern_Rivals_Page253.png)
Let AEB, GEH be two intersecting Lines, and CD another Line. It is to be proved that they cannot both be separational from CD.
In CD take any point F; and join EF.
Now, if possible, let AB, GH both be separational from CD;
∴ ∠s AEF, GEF are both equal to ∠EFD; [(α).
∴ they are equal to each other ; which is absurd.
Therefore two intersecting Lines &c. Q. E. D.