For, with the same construction,
since the rectangle AB, F is equal to the rectangle CD, E,
and the rectangle AB, F is BG,
for AG is equal to F,
and the rectangle CD, E is DH,
for CH is equal to E,
therefore BG is equal to DH.
And they are equiangular.
But in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. [VI. 14]
Therefore, as AB is to CD, so is CH to AG.
But CH is equal to E, and AG to F;
therefore, as AB is to CD, so is E to F.