To cut a given finite straight-line in half. Let $AB$ be the given finite straight-line. So it is required to cut the finite straight-line $AB$ in half.  Let the equilateral triangle $ABC$ have been constructed upon  ($AB$) [Prop.~1.1], and let the angle $ACB$ have been cut in half by the straight-line $CD$ [Prop.~1.9]. I say that the straight-line $AB$ has been cut in half at  point $D$.  For since $AC$ is equal to $CB$, and $CD$ (is) common, the two (straight-lines) $AC$, $CD$ are equal to the two (straight-lines) $BC$, $CD$, respectively. And the angle $ACD$ is equal to the angle $BCD$. Thus, the base $AD$ is equal to the base $BD$ [Prop.~1.4]. Thus, the given finite straight-line $AB$ has been cut in half at  (point) $D$.  (Which is) the very thing it was required to do.