Proof: By Euclid

Let the points $D$ and $E$ have been taken at random on each of the (straight lines) $CD$ and $CE$ (respectively), and let $DE$ have been joined.
And let the triangle $AFG$ have been constructed from three straight lines which are equal to $CD$, $DE$, and $CE$, such that $CD$ is equal to $AF$, $CE$ to $AG$, and further $DE$ to $FG$ [Prop. 1.22].
Therefore, since the two (straight lines) $DC$, $CE$ are equal to the two (straight lines) $FA$, $AG$, respectively, and the base $DE$ is equal to the base $FG$, the angle $DCE$ is thus equal to the angle $FAG$ [Prop. 1.8].
Thus, the rectilinear angle $FAG$, equal to the given rectilinear angle $DCE$, has been constructed at the (given) point $A$ on the given straight line $AB$.
(Which is) the very thing it was required to do.
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