To cut a given rectilinear angle in half.  Let $BAC$ be the given rectilinear angle. So it is required to cut it in half.  Let the point $D$ have been taken at random on $AB$, and let $AE$, equal to $AD$,  have been cut off from $AC$  [Prop.~1.3], and let $DE$ have been joined. And let the equilateral triangle $DEF$ have been constructed upon $DE$ [Prop.~1.1], and let $AF$ have been joined. I say that the angle $BAC$ has been cut in half by the straight-line $AF$.  For since $AD$ is equal to  $AE$, and $AF$ is common, the two (straight-lines) $DA$, $AF$ are equal to the two (straight-lines) $EA$, $AF$, respectively. And the base $DF$ is equal to the base $EF$. Thus, angle $DAF$ is equal to angle $EAF$ [Prop.~1.8].  Thus, the given rectilinear angle $BAC$ has been cut in half by the straight-line $AF$. (Which is) the very thing it was required to do.