Proof: By Euclid

For let $CE$ have been drawn through point $C$ parallel to the straight line $AB$ [Prop. 1.31].
And since $AB$ is parallel to $CE$, and $AC$ has fallen across them, the alternate angles $BAC$ and $ACE$ are equal to one another [Prop. 1.29].
Again, since $AB$ is parallel to $CE$, and the straight line $BD$ has fallen across them, the external angle $ECD$ is equal to the internal and opposite (angle) $ABC$ [Prop. 1.29].
But $ACE$ was also shown (to be) equal to $BAC$.
Thus, the whole angle $ACD$ is equal to the (sum of the) two internal and opposite (angles) $BAC$ and $ABC$.
Let $ACB$ have been added to both.
Thus, (the sum of) $ACD$ and $ACB$ is equal to the (sum of the) three (angles) $ABC$, $BCA$, and $CAB$.
But, (the sum of) $ACD$ and $ACB$ is equal to two right angles [Prop. 1.13].
Thus, (the sum of) $ACB$, $CBA$, and $CAB$ is also equal to two right angles.
Thus, in any triangle, (if) one of the sides (is) produced (then) the external angle is equal to the (sum of the) two internal and opposite (angles), and the (sum of the) three internal angles of the triangle is equal to two right angles.
(Which is) the very thing it was required to show.
∎

Thank you to the contributors under CC BY-SA 4.0!