To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.

Let $ABC$ be the given triangle, and $D$ the given rectilineal angle;thus it is required to construct in the recilineal angle $D$ a parallelogram equal to the triangle $ABC$.Let $BC$ be bisected at $E$, and let $AE$ be joined;on the straight line $EC$, and at the point $E$ on it, let the angle $CEF$ be constructed equal to the angle $D$; [Prop. 1.23]through $A$ let $AG$ be drawn parallel to $EC$, and [Prop. 1.31]through $C$ let $CG$ be drawn parallel to $EF$.Then $FECG$ is a parallelogram.And, since $BE$ is equal to $EC$,the triangle $ABE$ is also equal to the triangle $AEC$,for they are on equal bases $BE$, $EC$ and in the same parallels $BC$, $AG$; [Prop. 1.38]therefore the triangle $ABC$ is double of the triangle $AEC$.But the parallelogram $FECG$ is also double of the triangle $AEC$, for it has the same base with it and is in the same parallels with it; [Prop. 1.41]therefore the parallelogram $FECG$ is equal to the triangle $ABC$.And it has the angle $CEF$ equal to the given angle $D$.Therefore the parallelogram $FECG$ has been constructed equal to the given triangle $ABC$, in the angle $CEF$ which is equal to $D$.Q.E.F.