To construct a parallelogram equal to a given triangle in a given rectilinear angle. Let $ABC$ be the given triangle, and $D$ the given rectilinear angle. So it is required to construct a parallelogram equal to triangle $ABC$ in the rectilinear angle $D$.  <prf> 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 $D$.