If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole with the added straight line and the added straight line, together with the square on the half, is equal to the square on the straight line made up of the half and the added straight line.
Proof
Let AB be bisected at C (I.10) and produced to D, so that CD
is the half plus the added segment BD. Describe on CD the
square CEFD (I.46), join DE, and through B draw BG parallel
to CE or DF (I.31), meeting DE at H and EF at G.
Through H draw KM parallel to AD or EF (I.31), and through
A draw AK parallel to CL or DM (I.31).
As in II.5, the complement CH equals the complement HF (I.43).
Adding the square LHMG (which is the square on BC=CB) to both
of CH+AL (the rectangle on AD, DB) shows that the rectangle
AD⋅DB together with the square on CB equals the gnomon plus
the small square, which is the square CEFD on CD. Hence
AD⋅DB+CB2=CD2 as required.