Construction of Parallelogram Equal to Triangle in Given Angle
Bisect $BC$ at $E$, and join $AE$.
Construct $\angle CEF$ equal to $\angle D$.
Then $FEGC$ is a parallelogram.
Since $BE = EC$, from Triangles with Equal Base and Same Height have Equal Area, $\triangle ABE = \triangle AEC$.
So $\triangle ABC$ is twice the area of $\triangle AEC$.
But from Parallelogram on Same Base as Triangle has Twice its Area, $FECG$ is also twice the area of $\triangle AEC$.
So $FECG$ has the same area as $\triangle ABC$, and has the given angle $D$.