PROPOSITION 2.
If the square on a straight line be five times the square on a segment of it,
then,
when the double of the said segment is cut in extreme and mean ratio,
the greater segment is the remaining part of the original straight line.
For let the square on the straight line
AB be five times the square on the segment
AC of it, and let
CD be double of
AC; I say that, when
CD is cut in extreme and mean ratio, the greater segment is
CB.
Let the squares
AF,
CG be described on
AB,
CD respectively, let the figure in
AF be drawn, and let
BE be drawn through.
Now, since the square on
BA is five times the square on
AC,
AF is five times
AH.
Therefore the gnomon
MNO is quadruple of
AH.
And, since
DC is double of
CA, therefore the square on
DC is quadruple of the square on
CA, that is,
CG is quadruple of
AH.
But the gnomon
MNO was also proved quadruple of
AH; therefore the gnomon
MNO is equal to
CG.
And, since
DC is double of
CA, while
DC is equal to
CK, and
AC to
CH, therefore
KB is also double of
BH. [
VI. 1]
But
LH,
HB are also double of
HB; therefore
KB is equal to
LH,
HB.
But the whole gnomon
MNO was also proved equal to the whole
CG; therefore the remainder
HF is equal to
BG.
And
BG is the rectangle
CD,
DB, for
CD is equal to
DG; and
HF is the square on
CB; therefore the rectangle
CD,
DB is equal to the square on
CB.
Therefore, as
DC is to
CB, so is
CB to
BD.
But
DC is greater than
CB; therefore
CB is also greater than
BD.
Therefore, when the straight line
CD is cut in extreme and mean ratio,
CB is the greater segment.
Therefore etc. Q. E. D.
LEMMA.
That the double of
AC is greater than
BC is to be proved thus.
If not, let
BC be, if possible, double of
CA.
Therefore the square on
BC is quadruple of the square on
CA; therefore the squares on
BC,
CA are five times the square on
CA.
But, by hypothesis, the square on
BA is also five times the square on
CA; therefore the square on
BA is equal to the squares on
BC,
CA: which is impossible. [
II. 4]
Therefore
CB is not double of
AC.
Similarly we can prove that neither is a straight line less than
CB double of
CA; for the absurdity is much greater.
Therefore the double of
AC is greater than
CB. Q. E. D.