Proof: By Euclid

Let the (straight line) $AC$ have been cut in half at (point) $E$ [Prop. 1.10].
And $BE$ being joined, let it have been produced in a straight line to (point) $F$.¹ And let $EF$ be made equal to $BE$ [Prop. 1.3], and let $FC$ have been joined, and let $AC$ have been drawn through to (point) $G$.
Therefore, since $AE$ is equal to $EC$, and $BE$ to $EF$, the two (straight lines) $AE$, $EB$ are equal to the two (straight lines) $CE$, $EF$, respectively.
Also, angle $AEB$ is equal to angle $FEC$, for (they are) vertically opposite [Prop. 1.15].
Thus, the base $AB$ is equal to the base $FC$, and the triangle $ABE$ is equal to the triangle $FEC$, and the remaining angles subtended by the equal sides are equal to the corresponding remaining angles [Prop. 1.4].
Thus, $BAE$ is equal to $ECF$.
But $ECD$ is greater than $ECF$.
Thus, $ACD$ is greater than $BAE$.
Similarly, by having cut $BC$ in half, it can be shown (that) $BCG$ - that is to say, $ACD$ - (is) also greater than $ABC$.
Thus, for any triangle, when one of the sides is produced, the external angle is greater than each of the internal and opposite angles.
(Which is) the very thing it was required to show.

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The implicit assumption that the point $F$ lies in the interior of the angle $ABC$ should be counted as an additional postulate (translator's note) ↩