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TO make an Angle equal to an Angle given in any Point of a Line.
Let it be proposed to make an Angle equal to EDF, at the Point A, of the Line AB, at the Points A and D, as Centers, draw two Arches BC, EF, with the same extent of the Compasses; then take the Distance EF between your Compasses, put one Foot in B, and cut off BC, and draw AC. I say that the Angles BAC, EDF, are equal.
Demonstration. The Triangles ABC, DEF, have the Sides AB, AC equal to the Sides DE, DF; since that the Arches BC, EF, were described with the same extent of the Compass, they have also their Bases BC, EF, equal: Therefore the Angles BAC, DEF, are equal (by the 8th.)
THis Problem is so necessary in Sur∣veying Fortifications, Prospective, Dialling, and in all other parts of the Mathematicks; so that the greatest part