RESPECTING THE THEORY OF PARALLELS. 441 



Prop. XII. If straight lines he drawn to two 'parallel straight lines on the same side 

 of the common perpendicular, making with them equal angles respectively to- 

 wards the same sides, that which is nearer to the common perpendicular is less 

 than one more remote ; and two equal lines can he drawn, making equal angles 

 towards the same side with the given parallels, one on each side of the common 

 perpendicular. 



Let AB be parallel to CD (fig. Prop. X.), EF perpendicular to both ; GH, LM 

 straight lines which make the angles EGH = FHG and ELM=FML ; LM is 

 greater than GH. 



The angles HGL, GHM are each greater than a right angle, and GIL M, HML 

 each less than a right angle (Ax. Cor 2). Therefore -=^HGL = GLM. 



And because GL=HM (Prop. XI. Cor. 2), therefore (Euc. I. 4) HL = GM. 



Now if LM be not greater than GH, it is either equal to it or less than it. It 

 is not equal to it, for then the triangles HGL, GLM would be equal in ever}' 

 respect, and the -i::HGL = GLM, which it is not. 



Neither is LM^GH, for, if possible, produce ML to N, making MN = HG ; 

 join HN. Then, in the triangles LGH, HMN, the angle LGH is greater than 

 HMN, therefore (Euc. i. 24) HL is greater than HN, and .^HNL^HLN ; but HLN 

 is greater than a right angle, because HLM is less than a right angle, therefore HNL 

 is greater than a right angle, which is absurd : LM is therefore greater than GH. 



Next, if EP = EG, FQ=FH, PQ will be equal to GH, and will make equal 

 angles with the parallels towards the same side. 



For PF=FG, and ^PFE=EFG, therefore ^PFQ=GFH, and the triangle 

 PFQ= GFH. Therefore -^PQF=GHF and PQ= GH. 



Similarly QPE = HGE. 



Prop. XIII. The straight line which makes with two parallel straight lines the alter- 

 nate angles equal, is less than any other straight line which can he drawn from 

 one of the parallels, to make with the other an angle equal to either of these 

 alternate angles towards the same side. 



Let PQ make with the parallel straight lines AB, CD, the alternate angles 

 equal to one another ; let also any other straight line, 

 RS, make the angle RSD = PQD ; PQ is less than RS, '^ T/^ pf ^ 



Bisect PQ in K, draw KF perpendicular to CD. /p /\ 



make PE = QF, and join KE, then (Euc. I. 4 and 14) " ^ ^ ^~'' ° 



EKF is a straight line perpendicular to AB. Also, if SL be made =QK, and LH 



VOL. XXIII, PART III. 6 D 



