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 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 extrer mean ratio, CB is the greater segment.

Therefore etc.

LEMMA.

Q. E. 1

That the double of AC is greater than BC is to be f

thus.

If not, let BC be, if possible, double of CA. Therefore the square on BC is quadruple of the on CA;

therefore the squares on BC, CA are five times the s on CA.

But, by hypothesis, the square on BA is also five the square on CA;

therefore the square on BA is equal to the squares on BC which is impossible.

Therefore CB is not double of AC.

Similarly we can prove that neither is a straight lin

than CB double of CA;

for the absurdity is much greater.

Therefore the double of AC is greater than CB.

This proposition is the converse of Prop. 1. AB be so divided at C that

and if CD = 2AC, then

Q. E. I

We have to prove

(sq. on AB) = 5 (sq. on AC),

(rect. CD, DB) = (sq. on CB).

Subtracting these equals from the equals, the square on CD a gnomon MNO respectively, we have

i.e.

BG= (square HF),

(rect. CD, DB) = (sq. on CB).

Here again the proposition can readily be proved by means of a similar to that of II. II.

Draw CA through C at right angles to CB and of length equal to the original figure; make CD double of CA; produce AC to R so that CR = CB.

Complete the squares on CB and CD, and join AD.

Now we are given the fact that

(rect. KR, RC) + (sq. on AC) = (sq. on AR)

R

A

E

B

Heiberg, with reason, doubts the genuineness of the Lemma followin proposition.

PROPOSITION 3.

If a straight line be cut in extreme and mean rati square on the lesser segment added to the half of the gr segment is five times the square on the half of the gr segment.

446

BOOK XIII

(XIII. 3

For let any straight line AB be cut in extreme and mean ratio at the point C,

And, since the rectangle AB, BC is equal to the square on AC,

and CE is the rectangle AB, BC,

therefore CE is equal to RS.

But RS is quadruple of FG ; therefore CE is also quadruple of FG. Again, since AD is equal to DC,

HK is also equal to KF.

Hence the square GF is also equal to the square HL.
Therefore GK is equal to KL, that is, MN to NE;

hence MF is also equal to FE.

But MF is equal to CG; therefore CG is also equal to FE.

Let CN be added to each;

therefore the gnomon OPQ is equal to CE.

But CE was proved quadruple of GF;

therefore the gnomon OPQ is also quadruple of the square FG. Therefore the gnomon OPQ and the square FG are five times FG.

But the gnomon OPQ and the square FG are the square DN.

And DN is the square on DB, and GF the square on DC.
Therefore the square on DB is five times the square

on DC.

Q. E. D.

The theorem is still more obvious if the figure of II. I be used. Let CF be divided in extreme and mean ratio at E, by the method of II. II.

and AB is divided at C in extreme and mean ratio. (sq. on BD) = (sq. on DF)

And

= 5 (sq. on CD).

PROPOSITION 4.

E

If a straight line be cut in extreme and mean ratio, t square on the whole and the square on the lesser segment togeth are triple of the square on the greater segment.

Let AB be a straight line,

let it be cut in extreme and mean ratio at C, and let AC be the greater segment;

I say that the squares on AB, BC are triple of the square on CA.

For let the square ADEB be de

scribed on AB,

and let the figure be drawn.

Since then AB has been cut in extreme

and mean ratio at C,

and AC is the greater segment,

therefore the rectangle AB, BC is equal to the square on A

[VI. Def. 3, VI. And AK is the rectangle AB, BC, and HG the squ on AC:

therefore AK is equal to HG.

448

BOOK XIII

[XIII. 4, 5

And, since AF is equal to FE,

let CK be added to each;

therefore the whole AK is equal to the whole CE ; therefore AK, CE are double of AK.

But AK, CE are the gnomon LMN and the square CK; therefore the gnomon LMN and the square CK are double of AK.

But, further, AK was also proved equal to HG;

therefore the gnomon LMN and the squares CK, HG are triple of the square HG.

And the gnomon LMN and the squares CK, HG are the whole square AE and CK, which are the squares on AB, BC,

while HG is the square on AC.

Therefore the squares on AB, BC are triple of the square on AC.

Q. E. D.

Here, as in the preceding propositions, the results are proved de novo by the method of Book II., without reference to that Book. Otherwise the proof might have been shorter.

For, by II. 7,

(sq. on AB) + (sq. on BC) = 2 (rect. AB, BC) + (sq. on AC)

If a straight line be cut in extreme and mean ratio, and there be added to it a straight line equal to the greater segment, the whole straight line has been cut in extreme and mean ratio, and the original straight line is the greater segment.

For let the straight line AB be cut in extreme and mean ratio at the point C,

let AC be the greater segment, and let AD be equal to AC.

I say that the straight line DB has been cut in extreme and mean ratio at A, and the original straight line AB is the greater segment.

A

C

B

K

H

For let the square AE be described on AB, and let the figure be drawn.

E