Problem. To bisect a given rectilineal angle (BAC).
Sol.—In AB take any point D, and cut off [iii.] AE equal to AD. Join DE (Post. i.), and upon it, on the side remote from A, describe the equilateral triangle DEF [i.] Join AF. AF bisects the given angle BAC.
Dem.—The triangles DAF, EAF have the side AD equal to AE (const.) and AF common; therefore the two sides DA, AF are respectively equal to EA, AF, and the base DF is equal to the base EF, because they are the sides of an equilateral triangle (Def. xxi.). Therefore [viii.] the angle DAF is equal to the angle EAF; hence the angle BAC is bisected by the line AF.
Cor.—The line AF is an axis of symmetry of the figure.