Commentary on Proposition XIII.5

Back to Proposition XIII.5.

[Folio 0453 | p. 449]

We have

(sq. DH) = (sq. HC) = (rect. CE), by hypothesis, = (rect. HE).

Add to each side the rectangle AK, and

(rect. DK) = (sq. AE), or (rect. BD, DA) = (sq. on AB).

The result is of course obvious from II. 11.

There is an alternative proof given in P after XIII. 6, which depends on Book V.

By hypothesis,

BA : AC = AC : CB,

or, inversely,

AC : AB = CB : AC.

Componendo,

(AB + AC) : AB = AB : AC,

or

DB : BA = BA : AD.