To bisect a given finite straight line.

Let $AB$ be the given finite straight line.Thus it is required to bisect the finite straight line $AB$.Let the equilateral triangle $ABC$ be constructed on it, [Prop. 1.1]and let the angle $ACB$ be bisected by the straight line $CD$; [Prop. 1.9]I say that the straight line $AB$ has been bisected at the point $D$.For since $AC$ is equal to $CB$, and $CD$ is common,the two sides $AC$, $CD$ are equal to the two sides $BC$, $CD$ respectively;and the angle $ACD$ is equal to the angle $BCD$;therefore the base $AD$ is equal to the base $BD$. [Prop. 1.4]Therefore the given finite straight line $AB$ has been bisected at $D$.Q.E.F.