If two straight-lines cut one another then they make the vertically opposite angles equal to one another.  For let the two straight-lines $AB$ and $CD$ cut one another at the point $E$. I say that  angle $AEC$ is equal to (angle) $DEB$, and (angle) $CEB$ to (angle) $AED$.  For since the straight-line $AE$ stands on the straight-line $CD$, making the angles $CEA$ and $AED$, the (sum of the) angles $CEA$ and $AED$ is thus equal to two right-angles [Prop.~1.13]. Again, since the straight-line $DE$ stands on the straight-line $AB$, making the angles $AED$ and $DEB$, the (sum of the) angles $AED$ and $DEB$ is thus equal to two right-angles [Prop.~1.13]. But (the sum of) $CEA$ and $AED$ was also shown (to be) equal to two right-angles. Thus, (the sum of) $CEA$ and $AED$ is equal to (the sum of) $AED$ and $DEB$ [C.N.~1]. Let $AED$ have been subtracted from both. Thus, the remainder $CEA$ is equal to the remainder $BED$ [C.N.~3]. Similarly, it can be shown that $CEB$ and $DEA$ are also equal.  Thus, if two straight-lines cut one another then they make the vertically opposite angles equal to one another. (Which is) the very thing it was required to show.