# Proof: By Euclid

• 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.

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