Solution.

Construct the parallelogram B E F G equal to the given triangle C, and having the angle B equal to D, and so that B E be in the same right line with A B; and produce F G, and through A draw A H parallel to B G, and join H B. Then because H L and F K are parallel the angles L H F and F are together equal to two right angles, and therefore B H F and F are together less than two right angles, and therefore H B and F E being produced will meet as at K. Produce H A and G B to meet K L parallel to H F, and the parallelogram A M will be that which is required.

Demonstration.

It is evidently constructed on the given line A B; also in the parallelogram F L, the parallelograms A M and G E are equal (XLIII); but G E is equal to C (const.), therefore A M is equal to C. The angle E B G is equal to A B M (XV), but also to D (const.), therefore A B M is equal to D. Hence A M is the parallelogram required.