**Problem X:** Given three straight lines AG, BC and AD, to find a fourth proportional DE, that is so that AB is to BC as AD is to DE.

This problem can be done in the manner of Euclid, putting the first two lines AB, BC in a straight line AC, and the third on another line AE, which forms an angle CAE with AC. If BD is drawn parallel to CE, then DE is the fourth proportional being sought.