considered in a new Method. 



221 



each other. (This can be shown as a simple consequence of the 

 4th of Euchd's first book ; but is manifest more simply by 

 supposing wK^y to be folded over upon AB ; for, on account of 

 the right angles, wA, kx and ^/B, ^z would coincide throughout 

 their entire length.) 



Proposition. Lemma I. 



Let wx and yz be two parallel lines, 

 AB the line to which both are perpen- 

 dicular. Take D and E, equally distant 

 from A and B, on the same side of AB. 

 Bisect AB in C. Erect Cm perpendicular 

 to AB. Then shall Cu meet DE as in m, 

 a point which bisects DE ; and Cm shall 

 stand perpendicularly to DE. 



For Cm is parallel to both wx and yz, and 

 therefore, if produced, can never meet either 

 {Cor. I. Def.). Therefore if produced, it 

 must cross DJS between D and E. Suppose 

 at m, now join Am and Bm. Then in the 

 triangles ACm and BC?w we have the sides 



AC and Cm respectively equal to BC and Cm, and the angles 

 contained by these sides (both right) equal, therefore (4th, first 

 book of Euclid) Am = Bm, and the angles CmA=CmB, and 

 CAm = CBm; and since the angle DAC=EBC (both right), 

 hence mAD = mBE. Then in the triangles mAD and mBE we 

 have the sides DA and Am respectively equal to EB and Bm, 

 and the angles contained by those sides also equal ; therefore 

 (4th, Euclid's first book) Dm=Em ; and the angle DmA=EmA. 

 Therefore we have DmC = DmA + AmC = EmB + BmC = EmC, 

 and therefore mC perpendicular to DE, and bisects it. Q.E.D. 



Corollary, Hence also angle EDA = DEB. 



Proposition. Lemma II. 



Let AD and BE (fig. 1 or fig. 2) be any parallel and equally 

 long lines both perpendicular to AB. Join D and E. Then 

 shall the joining line stand perpendicularly both to AD and 

 to BE. 



Take four points, a, d, V, and Z>, the two first near to and 

 equally distant on each side from A ; the two latter, the same 

 distances on each side of B. Erect the four lines ad, dd\ 6V, 

 and he, each perpendicular to AB, and each equal in length to 

 AD or BE. Join d with ^ by the straight line dd\ and e' with 

 e by the straight line e^e. These joining lines are each respect- 

 ively at right angles to AD and BE (produced if necessary), and 

 are bisected by them {Lemma I.). At D, draw the straight line 



