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](elem.6.1)]

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](elem.2.4)]

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.