**Problem VIII:** Above a given straight line AB, to construct an equilateral triangle.

Construction: Cut AB by the second problem, into two equal parts at C, and from C, above AB, by the 5th problem, erect a perpendicular CF and locate on it DC equal to AC or CB. Draw DB and above that construct, as in the 5th problem, a perpendicular to DE at D, and equal to DC, and join EB. I say that if CF is made equal to BE and the triangle AF, FB is made, then the triangle AFB is equilateral.