440 PROFESSOR KELLAND ON THE LIMITS OF OUR KNOWLEDGE 



respect, and EH=KF; whence also the triangles EFK, EHK are equal in every 

 respect; and -=^HKE=FEK, and each is greater than a right angle, which 

 is impossible (Ax. Cor. 2.) HG is therefore greater than EF. 



Next if L be a point in EB beyond G, from which the perpendicular LM is 

 drawn to CD, LM is greater than GH. 



The angle LGH is greater than a right angle, and the angle GLM less ; there- 

 fore LGH:::^GLM. Now, we can prove, exactly as in the former case, that if 

 LM = GH, -^GLM=LGH, which is impossible; and if LM-^GH, by producing 

 ML to N, and joining GN, we can prove that -^GNM=NGH, or greater than a 

 right angle, which is impossible ; therefore LM is greater than GH. 



Lastly, Two equal perpendiculars can be drawn, one on each side of EF. 



Take EP=:EG and FQ = FH ; join PQ ; PQ is equal to GH, and perpendicular 

 to CF. 



For PF = FG and -::PrE = EFG (Euc. i. 4), therefore ^PFQ=GFH ; and 

 the triangle PFQ, is equal to the triangle GFH in every respect. Hence -^PQF 

 = GHF, and is a right angle ; and PQ = GH. 



Cor. The angles MLG, HGL are less than two right angles, and the angles 

 HGE, HGL are equal to two right angles ; therefore -^^MLG is less than HGE ; 

 that is, if lines be drawn from one of the parallels perpendicular to the other, 

 the interior angles in which they cut the first parallel continually diminish, as 

 the distances from the common perpendicular increase. 



Prop. XL If a straight line he perpendicular to each of two parallel straight 

 lines, and from the points at which it meets them equal distances he set off along 

 those lines towards the same side, the straight line which joins the points so 

 found shall make equal angles with each of the parallels towards the same side. 



In the figure of Prop. X., let EF be perpendicular to both the parallels AB, 

 CD; and let EG=FH, the angle EGH is equal to FHG. The triangles GEF, 

 HFE are equal, therefore GF = EH, and consequently the triangles EGH, FHG 

 are equal, and the angle EGH = FHG, 



Cor. 1. If the angles which a straight line makes with each of two parallels 

 towards the same side are equal, the straight line meets the parallels at points 

 which are equidistant from the points where the common perpendicular meets 

 them. 



Cor. 2. If two straight lines be drawn, each making equal angles with two 

 parallels towards the same side, the portions of each parallel which they cut off 

 shall be equal. 



