Let $ABC$ be a right-angled triangle having the (angle) $A$ a right angle. And let the perpendicular $AD$ have been drawn. * I say that the (rectangle contained) by $CBD$ is equal to the (square) on $BA$, and the (rectangle contained) by $BCD$ (is) equal to the (square) on $CA$, and the (rectangle contained) by $BD$ and $DC$ (is) equal to the (square) on $AD$, and, further, the (rectangle contained) by $BC$ and $AD$ [is] equal to the (rectangle contained) by $BA$ and $AC$.
(not yet contributed)
Proofs: 1