For let the square CDEB be described on CB; [I. 46]
let ED be drawn through to F, and through A let AF be drawn parallel to either CD or BE. [I. 31]
Then AE is equal to AD, CE.
Now AE is the rectangle contained by AB, BC,
for it is contained by AB, BE, and BE is equal to BC;
AD is the rectangle AC, CB, for DC is equal to CB;
and DB is the square on CB.
Therefore the rectangle contained by AB, BC is equal to the rectangle contained by AC, CB together with the square on BC.