Then, since, as BC is to GH, so is GH to CF,
and, if three straight lines be proportional,
as the first is to the third, so is the figure on the first to the similar and similarly situated figure described on the second, [VI. 19, Por.]
therefore, as BC is to CF, so is the triangle ABC to the triangle KGH.
But, as BC is to CF, so also is the parallelogram BE to the parallelogram EF. [VI. 1]
Therefore also, as the triangle ABC is to the triangle KGH, so is the parallelogram BE to the parallelogram EF;
therefore, alternately, as the triangle ABC is to the parallelogram BE, so is the triangle KGH to the parallelogram EF. [V. 16]
But the triangle ABC is equal to the parallelogram BE;
therefore the triangle KGH is also equal to the parallelogram EF.
But the parallelogram EF is equal to D;
therefore KGH is also equal to D.
And KGH is also similar to ABC.