ABC. Make by the preceding Problem the Parallelogram BFGH, having the Angle HBF equal to the Angle E, and which may be equal to the Triangle ABC. Continue the Sides GH, CF, and make HI equal to the Line D; then draw the Line IBN, and by the Points I and N, draw IL Parallel to GN, and NL Parallel to GI; and continue HB to M, and FB to K. The Parallelogram KLBM is the Parallelogram required.
Demonstration. The Angle HBF equal to the given Angle E, is also equal to the Angle KBM (by the 15th.) and the Side KB is equal to the Line HI or D. Lastly, the Parallelogram MK is equal (by the foregoing) to the Pa∣rallelogram GFBH; and this was made equal to the Triangle ABC. There∣fore the Parallelogram MK is equal to the Triangle ABC.
USE.
IN this Proposition is contained a sort of Geometrical Division: For in Arithmetical Division there is proposed a number which may be imagined to be like a Rectangle: For example, the Rectangle