If then AG is equal to C, that which was enjoined will have been done;
for there has been applied to the given straight line AB the parallelogram AG equal to the given rectilineal figure C and deficient by a parallelogrammic figure GB which is similar to D.
But, if not, let HE be greater than C.
Now HE is equal to GB;
therefore GB is also greater than C.
Let KLMN be constructed at once equal to the excess by which GB is greater than C and similar and similarly situated to D. [VI. 25]
But D is similar to GB;
therefore KM is also similar to GB. [VI. 21]
Let, then, KL correspond to GE, and LM to GF.
Now, since GB is equal to C, KM,
therefore GB is greater than KM;
therefore also GE is greater than KL, and GF than LM.
Let GO be made equal to KL, and GP equal to LM; and let the parallelogram OGPQ be completed;
therefore it is equal and similar to KM.
Therefore GQ is also similar to GB; [VI. 21]
therefore GQ is about the same diameter with GB. [VI. 26]
Let GQB be their diameter, and let the figure be described.