To draw a straight-line perpendicular to a given infinite straight-line from a given point which is not on it.   Let $AB$ be the given infinite straight-line  and $C$ the given point, which is not on ($AB$). So it is required to draw a  straight-line  perpendicular to the given infinite straight-line $AB$ from the given point $C$, which is not on ($AB$).  For let point $D$ have been taken at random on the other side (to $C$) of  the straight-line $AB$, and let the circle $EFG$ have been drawn with center $C$ and radius $CD$ [Post.~3], and let the straight-line $EG$ have been cut in half at (point) $H$ [Prop.~1.10], and let the straight-lines $CG$, $CH$, and $CE$ have been joined. I say that the  (straight-line) $CH$ has been drawn  perpendicular to the given infinite straight-line $AB$ from the given point $C$, which is not on ($AB$).  For since $GH$ is equal to $HE$, and $HC$ (is) common, the two (straight-lines) $GH$,  $HC$ are equal to the two (straight-lines) $EH$, $HC$, respectively, and the base $CG$ is equal to the base $CE$. Thus, the angle $CHG$ is equal to the angle $EHC$ [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, and the former straight-line is called a perpendicular to that upon which it stands [Def.~1.10].  Thus, the (straight-line) $CH$ has been drawn perpendicular to the given infinite straight-line $AB$ from the given point $C$, which is not on  ($AB$). (Which is) the very thing it was required to do.