Let AB, CD be the given Pair of Lines.
Through the given point draw a Line perpendicular to AB, and let it meet AB in L. In AB take any 2 points E, F, equidistant from L. From E, F, draw EG, FH, perpendicular to CD. Bisect GH at K; and join KL.
Now E, F are 2 points, in AB, equidistant from CD; and GH is bisected in K, and EF in L;
∴ KL is a common perpendicular; [Lemma 1.
∴ it coincides with the Line drawn, through the given point, perpendicular to AB, since both meet AB at L;
∴ KL is the Line required.
Q. E. F.
(θ). II. 9.
A Pair of Lines, of which one has two points on the same side of, and equidistant from, the other, are equally inclined to any transversal.
![](http://upload.wikimedia.org/wikipedia/commons/thumb/0/0d/Euclid_and_His_Modern_Rivals_Page257.png/250px-Euclid_and_His_Modern_Rivals_Page257.png)
Let AB contain two points equidistant from CD, and let EF be a certain transversal: it shall be proved that ∠AEF = ∠EFD.
Now AB, CD, are equidistantial from each other. [(η).
Bisect EF at G; through G let HGK be drawn a common perpendicular to AB and CD. [Lemma 2.
Hence, in Triangles GEH, GFK, side GE and ∠s EGH, GHE, are respectively equal to side GF and ∠s FGK, GKF;
∴ ∠GEH = GFK. [Euc. I. 26.
Therefore a Pair of Lines, &c. Q. E. D.
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