** 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.*

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] [p. 444]

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.