To draw a straight-line at right-angles to a given straight-line from a given point on it.  Let $AB$ be the given straight-line, and $C$ the given point on it. So it is required to draw a straight-line from the point $C$ at right-angles to the straight-line $AB$.  Let the point $D$ be have been taken at random on $AC$, and let $CE$ be made equal to $CD$ [Prop.~1.3], and let the equilateral triangle $FDE$ have been constructed on $DE$ [Prop.~1.1], and let $FC$ have been joined. I say that the straight-line $FC$ has been drawn at right-angles to the given straight-line $AB$ from the given point $C$ on it.  For since $DC$ is equal to $CE$, and $CF$ is common, the two (straight-lines) $DC$, $CF$ are equal to the two (straight-lines), $EC$, $CF$, respectively.  And the base $DF$ is equal to the base $FE$. Thus, the angle $DCF$ is equal to the angle $ECF$ [Prop.~1.8], and they are adjacent. But when a straight-line stood on  a(nother) straight-line makes the adjacent angles equal to one another, each of the equal angles is a right-angle [Def.~1.10]. Thus, each of the (angles) $DCF$ and $FCE$ is a right-angle.  Thus, the straight-line $CF$ has been drawn at right-angles to the given straight-line $AB$ from the given point $C$ on it. (Which is) the very thing it was required to do.