# Proof: By Euclid

• In fact, if $CBA$ is equal to $ABD$ then they are two right angles [Def. 1.10] .
• But, if not, let $BE$ have been drawn from the point $B$ at right angles to [the straight line $CD$ [Prop. 1.11].
• Thus, $CBE$ and $EBD$ are two right angles.
• And since $CBE$ is equal to the two (angles) $CBA$ and $ABE$, let $EBD$ have been added to both.
• Thus, the (sum of the angles) $CBE$ and $EBD$ is equal to the (sum of the) three (angles) $CBA$, $ABE$, and $EBD$ [C.N. 2] .
• Again, since $DBA$ is equal to the two (angles) $DBE$ and $EBA$, let $ABC$ have been added to both.
• Thus, the (sum of the angles) $DBA$ and $ABC$ is equal to the (sum of the) three (angles) $DBE$, $EBA$, and $ABC$ [C.N. 2] .
• But (the sum of) $CBE$ and $EBD$ was also shown (to be) equal to the (sum of the) same three (angles).
• And things equal to the same thing are also equal to one another [C.N. 1] .
• Therefore, (the sum of) $CBE$ and $EBD$ is also equal to (the sum of) $DBA$ and $ABC$.
• But, (the sum of) $CBE$ and $EBD$ is two right angles.
• Thus, (the sum of) $ABD$ and $ABC$ is also equal to two right angles.
• Thus, if a straight line stood on a(nother) straight line makes angles, it will certainly either make two right angles, or (angles whose sum is) equal to two right angles.
• (Which is) the very thing it was required to show.

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