considered in a new Method. 



.223 



it (for a similar reason) ; and therefore Dm' must always lie 

 between wx and dd\ and between vJx^ and ee^, if they do not all 

 coincide, let the points a, a , b', and b be taken closer to A and B, or 

 further from them. But smcealwdLys ad = AD = a' d' = b'ef==SE=be, 

 we may make m' approach unlimitedly close to Eby takingi'B =Bd 

 unlimitedly small ; and similarly m will approach unlimitedly to 

 D, and e'e to rn'of, and d'd to wx ; and therefore in the limit they 

 all coincide ; and Dm', which in the limit coincides with DE, is 

 perpendicular to AD ; and therefore DE (which never changes 

 its position) is always perpendicular to AD ; and similarly (or by 

 corollary to Lemma I.) to BE. Q.E.D. 



Cor. I. Hence any line which is perpendicular 

 to one of two parallel lines AD, must, if pro- 

 duced, meet the other, and be perpendicular to 

 it also. For if a line be perpendicular at any 

 point D to AD, take BE=AD and join DE; 

 this joining line, being perpendicular to both, 

 must coincide with the perpendicular to AD 

 atD. 



Cor. 11. Hence all common perpendiculars 

 to two parallel lines meet them at equal di- 

 stances from A and B. 



Cor. III. Hence all common perpendiculars to two parallel 

 lines must be equal to one another throughout, and each must 

 be equal to AB. For since DE (any joining line) and AB are 

 both perpendicular to AD, they also are parallel ; and since BE 

 is perpendicular to both, AB=DE. 



Prop. 27 of Euclid's First Book, 



If a line (xy) falling on two other lines {wx and yz) make the 

 alternate angles equal {wxy — zyx)y these lines shall be parallel. 



For bisect xy in C, and let fall CA perpendicular on wXj and 

 CB perpendicular on yz. Then in the triangles xAG and ?/BC, 

 ^C = yC, and the angle Qxh. = CyB (hypothesis) and angle 

 a?AC = 2/BC (right); therefore (26th of Euclid's first book) the 

 angle a?CA=?/CB. But a?CA + ACy = two right angles (13th 

 of Euclid's first book) = BCy-f i/CA. Therefore AC and CB 

 are one continued line, and perpendicular to both wx and yz, 

 which are therefore parallel. Q.E.D. 



