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 rectilineal 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; [I. 23] through A let AG be drawn parallel to EC, and [I. 31] through C let CG be drawn parallel to EF. Then FECG is a parallelogram. And, since BE is equal to EC,