RESPECTING THE THEORY OF PARALLELS. 



489 



For ^ADB^DAE + DEA ^2AED, because AD = DE : that is, AED^ h ADB. 



Similarly, AFB^|AEB-^:|;ADB ; and it is evident that by proceeding in 

 the same way a line may be drawn from A to BC which shall make with BC an 

 angle less than any assignable angle. 



Prop. IX. Through a given 2'>oint without a straight line there may he drawn 

 infinite number of straight lines all jyarallel to the given sti'aight line. 



an 



Let A be the given point, BC the given straight line ; from the point A there 

 can be drawn an infinite number of straight lines all parallel to BC. 



Draw xiB perpendicular to BC, and AD perpendicular to AB. Take any 

 number of points E, G, &c., in BC, and join AE, 

 AG, &c. 



The angles BEA and BAE are less than a right 

 angle ; but the angles DAE and BAE are equal to 

 a right angle; therefore DAE^AEB. Cut off FAE = AEB. Again, FAE= FAG 

 and GAE ; but FAE = AEB ;^AGE and GAE. Take away GAE, and FAG:p-AGE. 

 Cut off HAG = AGE, &c. The straight lines AD, AF, AH, are all parallel to BC 

 (Euc. I. 27), and it is evident that their number is unlimited. 



Prop. X. I/a straight line be perpendicular to each of two parallel straight lines, 

 it is the least distance between them ; and of all other perpendiculars drawn 

 from, one of the parallels to the other, that which is nearer to the least is less 

 than one more rermte ; also two equal perpendiculars can be drawn, one on 

 each side of the shortest line. 



Let AB be parallel to CD; and let EF be perpendicular to both; EF is the 

 shortest distance between them. 



From G, any point in AB, draw GH perpendicular to CD : GH, and therefore 

 (Euc. L 19) any line drawn from G to CD is greater 

 than EF. 



A- 



If GH is not greater than EF, it is either equal to it 

 or less than it. c- 



First, Let, if possible, GH = EF; the triangles EFH, 

 GHF, are equal in every respect; therefore EH=GF; and the triangles EFG, 

 EHG, are equal in every respect; therefore .=:::EGH = GEE, and each is aright 

 angle, which is impossible (Ax. Cor. 2). 



Secondly, Let GH-=^EF, and produce it to K, so that HK = EF. As in the 

 former case, the triangles EFH, KHF (if KF be joined) are equal in every 



