Commentary on Proposition XIII.2

Back to Proposition XIII.2.

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.

This proposition is the converse of Prop. 1. We have to prove that, if AB be so divided at C that (sq. on AB) = 5 (sq. on AC), and if CD = 2 AC, then (rect. CD, DB) = (sq. on CB).

Subtract from each side the sq. on AC; then

(gnomon MNO) = 4 (sq. on AC) = (sq. on CD).

Now, as in the last proposition,

□ CE = 2 (□ BH) = □ BH + □ HL.

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

□ BG = (square HF),

i.e.

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

Here again the proposition can readily be proved by means of a figure similar to that of II. 11.

Draw CA through C at right angles to CB and of length equal to CA in 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.

R C B D A K E G

Now we are given the fact that

(sq. on AR) = 5 (sq. on CA).

But

5 (sq. on AC) = (sq. on AC) + (sq. on CD) = (sq. on AD).

Therefore

(sq. on AR) = (sq. on AD),

or AR $= $ AD.

Now

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

Therefore

(rect. KR . RC) = (sq. on CD).

That is,

(rectangle RE) = (square CG).

Subtract the common part CE, and

(rect. BG) = (sq. RB),

or

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

Heiberg, with reason, doubts the genuineness of the Lemma following this proposition.