Page:Carroll - Euclid and His Modern Rivals.djvu/295

Rh 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;

∴ 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.

. 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.



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.

Bisect EF at G; through G let HGK be drawn a common perpendicular to AB and CD.

Hence, in Triangles GEH, GFK, side GE and ∠s EGH, GHE, are respectively equal to side GF and ∠s FGK, GKF;

∴ ∠GEH = GFK.

Therefore a Pair of Lines, &c.