Now BC is a square;
therefore AD is also a square.
And, since BC is equal to CD,
let CE be subtracted from each;
therefore the remainder BF is equal to the remainder AD.
But it is also equiangular with it;
therefore in BF, AD the sides about the equal angles are reciprocally proportional; [VI. 14]
therefore, as FE is to ED, so is AE to EB.
But FE is equal to AB, and ED to AE.
Therefore, as BA is to AE, so is AE to EB.
And AB is greater than AE;
therefore AE is also greater than EB.
Complete text
Book I