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 $D$. 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.