PROPOSITION 6. THEOREM.
If a straight line he bisected, and produced to any point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced.
Let the straight line AB be bisected at the point C, and produced to the point D : the rectangle AD, DB, together with the square on CB, shall be equal to the square on CD.
On CD describe the
square CEFD ; [I. 46.
join DE through B draw
BHG parallel to CE or
DF; through H draw
KLM parallel to AD or
EF ; and through A draw
AK parallel to CL or DM. [I. 31.
Then, because AC equal to CB, [Hypothesis.
the rectangle AL is equal to the rectangle CH; [I. 36.
but CH is equal to HF; [I. 43.
therefore also AL is equal to HF. [Axiom 1.
To each of these add CM ;
therefore the whole AM is equal to the gnomon CMG. [Ax. 2.
But AM is, the rectangle contained by AD, DB,
for DM is equal to DB. [II. 4, Corollary.
Therefore the rectangle AD, DB is equal to the gnomon
CMG. [Axiom 1.
To each of these add LG, which is equal to the square on
CB. [II. 4, Corollary, and I. 34.
Therefore the rectangle AD, DB, together with the square
on CB, is equal to the gnomon CMG and the figure LG.
But the gnomon CMG and LG make up the whole figure
CEFD, which is the square on CD.
Therefore the rectangle AD,DB, together with the square
on CB, is equal to the square on CD.
Wherefore, if a straight line &c. q.e.d.
