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TO find a fourth Proportional to three Lines given.
Let there be proposed three Lines AB, BC, DE, to which must be found a fourth proportional, make an Angle as FAC, at discretion; take on AC the Lines AB, BC; and on AF, the Line AD equal to DE: then draw DB, and its parallel FC. I say that DF is the Line you seek for; that is to say, that there is the same Reason of AB to BC, as of DE, or AD to DF.
Demonstration. In the Triangle FAC, the Line DB is parallel to the Base FC; there is thence the same reason of AB to BC, as of AD to DF (by the 2d.)
THe use of the Compass of Proportion (or Sector) is established on these Propositions: for we divide a Line as we