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.
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.
