Solution.

Through the point A draw the right line A F parallel to B C, bisect B C the base of the triangle in E, and at the point E, and with the right line C E make the angle C E F equal to the given one D; through C draw C G parallel to E F until it meet the line A F in G. C F is the required parallelogram.

Demonstration.

Because E C is parallel to A G (const.), and E F parallel to C G, E G is a parallelogram, and has the angle C E F equal to the given one D (const.); and it is equal to the triangle B A C, because it is between the same parallels and on half of the base of the triangle (192).