INVESTIGATION of PORISMS. 193 



LEMMA I. Fig. io. 



32. If two ftraight lines, AE and BF, be cut by three other 

 ftraight lines, AB, CD and EF, given in pofition, and not 

 all parallel to one another, into fegments having the fame 

 given ratio, they will intercept between them fegments of 

 the lines given in pofition, viz. AB, CD, EF, which will 

 alfo have given ratios to one another *. 



PROP. 



* Demonstration. Through C and E draw CH and EG, both parallel to AB, 



and let them meet BG, parallel to AE, in H and in G. Let GF and HD be joined ; and 

 becaufe AC is to CE, that is, BH to HG as BD to DF, by hypothefis, DH is parallel to 

 GF, and has alfo a given ratio to it, vi%. the ratio of GB to BH, or of EA to AC. 

 Take GK equal to HD, and join EK, and the triangle EGK will be equal to the triangle 

 CHD, and therefore the angle KEG is given, and likewife the angle-KEF ; and fince 

 the ratio of GK to KF is given, if from K there be drawn KL parallel to EG, meeting 

 EF in L, the ratio of EL to LF will be given. But the ratio of EL to LK is 

 given, becaufe the triangle ELK is given in fpecies ; therefore the ratio of FL to LK is 

 given ; and the angle FLK being alfo given, the triangle FKL is given in fpecies, as alfo 

 the triangle FGE. The angle FGE being therefore given, the triangle KGE is given 

 in fpecies, and EG has therefore given ratios to EK and EF. But EG is equal to AB, 

 and EK to CD, therefore AB, CD and EF have given ratios to one another Q^E. D. 



Hence to find the ratios of AB, CD and EF ; in EF take any part EL, and make 

 as AC is to CE, fo EL to LF ; through L draw LK parallel to EG or AB, meeting 

 EK, drawn through E parallel to CD in K ; then if FK be drawn meeting EG in G, 

 the ratios required are the fame with the ratios of the lines EG, EK, EF. This is evi- 

 dent from the preceding inveftigation. 



If it be required to find the pofition of the line AE, drawn through the point A, fo as 

 to be cut by CD and EF in a given ratio ; draw Ac, any how, cutting DC in c, and pro- 

 duce Ac to e, fo that Ac may be to ce in the ratio which AC is to have to CE ; leteE 

 be drawn parallel to DC, interfering FE in E, and if AE be joined, it is the line re- 

 quired. 



Hence the converfe of the lemma is eafily demonflrated, viz. that if AE and BF be 

 two lines that are cut proportionally by the three lines AB, CD, EF ; and if AB and 

 EF, the parts of any two of thefe hill, intercepted between AE and BF, be alfo cut pro- 

 portionally, any how, in b and/, and if bf be joined, meeting the third line in d, bfWiW 

 be cut in the fame proportion with AE or BF. For if not, let bf be drawn from b, 

 meeting CD in d', and EF in/"', fo that bd' :d'f .: AC:CJL j then by the lemma, ab-.AB:: 

 E/:EF 5 and by fuppofition, a£:AB::E/:EF, therefore E/' = E/J which is impoffible. 

 Therefore, <&c. 



Vol. III. B b 



