To cut a given uncut straight line similarly to a given cut straight line.

Let *AB* be the given uncut straight line, and *AC* the straight line cut at the points *D* and *E,* and let them be so placed as to contain any angle.

Join *CB,* and draw *DF* and *EG* through *D* and *E* parallel to *CB,* and draw *DHK* through *D* parallel to *AB.*

Therefore each of the figures *FH* and *HB* is a parallelogram. Therefore *DH* equals *FG* and *HK* equals *GB.*

Now, since the straight line *EH* is parallel to a side *CK* of the triangle *DCK,* therefore, proportionally, *DE* is to *EC* as *DH* is to *HK.*

But *DH* equals *FG,* and *HK* equals *GB,* therefore *DE* is to *EC* as *FG* is to *GB.*

Again, since *DF* is parallel to a side *EG* of the triangle *AEG,* therefore, proportionally, *AD* is to *DE* as *AF* is to *FG.*

But it was also proved that *DE* is to *EC* as *FG* is to *GB,* therefore *DE* is to *EC* as *FG* is to *GB,* and *AD* is to *DE* as *AF* is to *FG.*

Therefore the given uncut straight line *AB* has been cut similarly to the given cut straight line *AC.*

Q.E.F.