Euclid, Elements
Thomas L. Heath, Sir Thomas Little Heath, Ed.

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thus it is required to apply to the given straight line AB a parallelogram equal to the given rectilineal figure C and deficient by a parallelogrammic figure which is similar to D.

Let AB be bisected at the point E, and on EB let EBFG be described similar and similarly situated to D; [VI. 18] let the parallelogram AG be completed.

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]