.
Sol.—Let it be required, for instance, to cut off the fourth part. Draw AF, making any angle with AB, and in AF take any point C, and cut off (I. iii.) the parts CD, DE, EF each equal to AC. Join BF, and draw CG parallel to BF. AG is the fourth part of AB.
Dem.—Since CG is parallel to the side BF of the triangle ABF, AC : AF :: AG : AB [ii.]; but AC is the fourth part of AF (const.). Hence AG is the fourth part of AB [V., d.]. In the same manner, any other required submultiple may be cut off.
Proposition x., Book I., is a particular case of this Proposition.
PROP. X.—Problem.
To divide a given undivided line (AB) similarly to a given divided line (CD).
Sol.—Draw AG, making any angle with AB, and cut off the parts AH, HI, IG respectively equal to the parts CE, EF, FD of the given divided line CD. Join BG, and draw HK, IL, each parallel to BG. AB will be divided similarly to CD.
Dem.—Through H draw HN parallel to AB, cutting IL in M. Now in the triangle ALI, HK is parallel to IL. Hence [ii.] AK : KL :: AH : HI, that is :: CE : EF (const.). Again, in the triangle HNG, MI is parallel to NG. Therefore [ii.] HM : MN :: HI : IG; but [I. xxxiv.] HM is equal to KL, MN is equal to LB, HI is equal to EF, and IG is equal to FD (const.). Therefore KL : LB :: EF : FD. Hence the line AB is divided similarly to the line CD.
Exercises.