If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half.
Proof
Let AB be bisected at C (I.10) and cut unequally at D.
Describe on CB the square CEFB (I.46), join BE, and through D
draw DG parallel to CE or BF (I.31), meeting BE at H and
EF at G. Through H draw KM parallel to AB or EF (I.31),
meeting CE at L and BF at M. Through A draw AK parallel
to CL or BM (I.31), meeting KM extended at K.
The complement CH equals the complement HF in the square CEFB
(I.43). Add to each the square DM; then the rectangle CDHL plus
the square LHMG equals the rectangle DBFG plus the same square.
But CDHL together with rectangle AC-equivalent piece AKLC
(which equals CDHL since AC=CB and the lines are parallel)
fills the gnomon NOP, plus the square LHMG on CD, equals the
square CEFB on CB. Thus the rectangle AD⋅DB together
with the square on CD equals the square on CB.