| 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. | I.31 | |
| Therefore each of the figures FH and HB is a parallelogram. Therefore DH equals FG and HK equals GB. | I.34 | |
| 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. | VI.2 | |
| But DH equals FG, and HK equals GB, therefore DE is to EC as FG is to GB. | V.7 | |
| Again, since DF is parallel to a side EG of the triangle AEG, therefore, proportionally, AD is to DE as AF is to FG. | VI.2 | |
| 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. | ||
Book VI Introduction - Proposition VI.9 - Proposition VI.11.