If from a parallelogram a(nother) parallelogram is subtracted (which is) similar, and similarly laid out, to the whole, having a common angle with it, then (the subtracted parallelogram) is about the same diagonal as the whole. * For, from parallelogram $ABCD$, let (parallelogram) $AF$ have been subtracted (which is) similar, and similarly laid out, to $ABCD$, having the common angle $DAB$ with it. * I say that $ABCD$ is about the same diagonal as $AF$.
If from a parallelogram ($ABCD$) a similar parellelogram ($AF$) is subtracted and both parallelograms share a common angle ($\angle{GAE}$), then their diagonals going through this angle lie on the same straight line.
Proofs: 1
