It is likewise rectangular. For since CG is parallel to
BK, and CB meets them, the angles KBC, GCB are together equal to two right angles. [I. 29.
But KBC is a right angle. [I. Definition 30.
Therefore GCB is a right angle. [Axiom 3.
And therefore also the angles CGK, GKB opposite to
these are right angles. [I. 34. and Axiom 1.
Therefore CGKB is rectangular ;
and it has been shewn to be equi-
lateral ; therefore it is a square, and
it is on the side CB.
For the same reason HF is also a
square, and it is on the side HG,
which is equal to AC. [I. 34.
Therefore HF, CK are the squares
on AC,CB.
And because the complement AG is equal to the complement GE; [I.43.
and that AG is the rectangle contained by AC, CB, for
CG is equal to CB;
therefore GE is also equal to the rectangle AC, CB. [Ax. 1.
Therefore AG, GE are equal to twice the rectangle AC, CB.
And HF, CK are the squares on AC, CB.
Therefore the four figures HF, CK, AG, GE are equal to
the squares on AC, CB, together with twice the rectangle
AC, CB.
But HF, CK, AG, GE make up the whole figure ADEB,
which is the square on AB.
Therefore the square on AB is equal to the squares on
AC, CB, together with twice the rectangle AC, CB.
Wherefore, if a straight line &c. q.e.d.
Corollary. From the demonstration it is manifest, that parallellograms about the diameter of a square are likewise squares.
PROPOSITION 5. THEOREM.
If a straight line be divided into two equal parts and also into two unequal parts, the rectangle contained by the
