incommensurable with A.

I. Euclid takes three straight lines of the form p, p √λ, p√ takes the mean proportional pλ between the first two

and the straight lines pλ‡, μà1 √1 – k2 satisfy the given conditions.

Now (a) pλt is medial.

(B) We have, from (1) and (2),

whence x ~ pλ‡; and x is therefore medial and ~

(y) x. pλ = p√√λ.p√1-k2.

ρ

But the latter is medial;

therefore x. pλ, or pλ. pλ √1 – k2, is medial.

Ρ I

Lastly (8) p, p√1 – k2 have the remaining property in the enun therefore pì, pλ √īk2 have it also.

(Euclid has not the assistance of symbols to prove the proportion He therefore uses the lemmas ab: bca: c and d2: de de to de the relations

and

that

The straight lines pλ‡, pλ

=

√1-k2 may take any of the follow according as the straight lines first taken are

(1) a, √B, √a2-c2, (2) √A, √B, JA-k2A, (3) √A, b, ↓

74

BOOK X

[x. 32, Lemma

II. If the other conditions are the same, but the square on the first medial straight line is to exceed the square on the second by the square on a straight line incommensurable with the first, we begin with the three straight ρ and the medial straight lines are

lines p, P√

The possible forms are even more various in this case owing to the more various forms that the original lines may take, e.g.

Let ABC be a right-angled triangle having the angle A right, and let the perpendicular AD be

drawn ;

I say that the rectangle CB, BD is equal to the square on BA,

the rectangle BC, CD equal to the square on CA,

B

A

the rectangle BD, DC equal to the square on AD,

and, further, the rectangle BC, AD equal to the rectangle BA, AC.

And since the triangle ABC is similar to the triang therefore, as CB is to BA, so is BA to BD; therefore the rectangle CB, BD is equal to the squar

For the same reason the rectangle BC, CD is al to the square on AC.

And since, if in a right-angled triangle a perpe be drawn from the right angle to the base, the perp so drawn is a mean proportional between the segmen base,

[

therefore, as BD is to DA, so is AD to DC; therefore the rectangle BD, DC is equal to the square

I say that the rectangle BC, AD is also equal to angle BA, AC.

For since, as we said, ABC is similar to ABD, therefore, as BC is to CA, so is BA to AD.

Therefore the rectangle BC, AD is equal to the BA, AC.

PROPOSITION 33.

Q. E

To find two straight lines incommensurable in squa make the sum of the squares on them rational but the contained by them medial.

Let there be set out two rational straight lines commensurable in square only and such that the square on the greater AB is greater than the square on the less BC by the square on a straight line incommensurable with AB,

[x. 30]

F

A

E B

let EF be drawn at right angles to AB, and let AF, FB be joined.

Then, since AB, BC are unequal straight lines, and the square on AB is greater than the square on the square on a straight line incommensurable with A while there has been applied to AB a parallelogram the fourth part of the square on BC, that is, to the so half of it, and deficient by a square figure, making t angle AE, EB,

therefore AE is incommensurable with EB.

And, as AE is to EB, so is the rectangle BA, A rectangle AB, BE,

while the rectangle BA, AE is equal to the square o and the rectangle AB, BE to the square on BF; therefore the square on AF is incommensurable square on FB;

therefore AF, FB are incommensurable in square. And, since AB is rational,

therefore the square on AB is also rational;

so that the sum of the squares on AF, FB is also ra

And since, again, the rectangle AE, EB is equ square on EF,

and, by hypothesis, the rectangle AE, EB is also equ square on BD,

therefore FE is equal to BD;

therefore BC is double of FE,

so that the rectangle AB, BC is also commensurable rectangle AB, EF.

But the rectangle AB, BC is medial ;

therefore the rectangle AB, EF is also medial.

in square have been found which make the sun squares on them rational, but the rectangle contained medial.

Euclid takes the straight lines found in x. 30, viz. p,

He then solves geometrically the equations

4(1+2)

If x, y are the values found, he takes u, v such that

u2 = px

v2 = py

√1 + k2

and u, v are straight lines satisfying the conditions of the problem. Solving algebraically, we get (if x>y)

Euclid's proof that these straight lines fulfil the requirements i

(a) The constants in the equations (1) satisfy the conditions of