If a parallelogram, falling short^{1} by a square figure, is applied to some straight line then the applied (parallelogram) is equal (in area) to the (rectangle contained) by the pieces of the straight line created via the application (of the parallelogram). * For let the parallelogram $AD$, falling short by the square figure $DB$, have been applied to the straight line $AB$. * I say that $AD$ is equal to the (rectangle contained) by $AC$ and $CB$.
(not yet contributed)
Proofs: 1
Proofs: 1
Note that this lemma only applies to rectangular parallelograms (translator's note). ↩