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FRom a Point C in a Right Line given AB, to erect a Right Line CF, at Right Angles.
Take on either side of this Point given two equal Lines, CD, CE, and upon the Right Line DE, erect an Equi∣lateral Triangle; draw the Line FC, and it will be the Perpendicular re∣quired.
Demonstration. The Triangles DCF, ECF, have the Side CF common, the Sides DF, EF, equal, the Bases DC, EC, are also equal: Therefore (by the 8th.) the Angles DCF, ECF, are
equal, (and by the 10th Def.) the Line CF is Perpendicular to AB.