Proof: By Euclid

Let $BC$ have been cut in half at $E$ [Prop. 1.10], and let $AE$ have been joined.
And let (angle) $CEF$, equal to angle $D$, have been constructed at the point $E$ on the straight line $EC$ [Prop. 1.23].
And let $AG$ have been drawn through $A$ parallel to $EC$ [Prop. 1.31], and let $CG$ have been drawn through $C$ parallel to $EF$ [Prop. 1.31].
Thus, $FECG$ is a parallelogram.
And since $BE$ is equal to $EC$, triangle $ABE$ is also equal to triangle $AEC$.
For they are on the equal bases, $BE$ and $EC$, and between the same parallels, $BC$ and $AG$ [Prop. 1.38].
Thus, triangle $ABC$ is double (the area) of triangle $AEC$.
And parallelogram $FECG$ is also double (the area) of triangle $AEC$.
For it has the same base as ($AEC$), and is between the same parallels as ($AEC$) [Prop. 1.41].
Thus, parallelogram $FECG$ is equal to triangle $ABC$.
($FECG$) also has the angle $CEF$ equal to the given (angle) $D$.
Thus, parallelogram $FECG$, equal to the given triangle $ABC$, has been constructed in the angle $CEF$, which is equal to $D$.
(Which is) the very thing it was required to do.
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