Commentary on Proposition III.15¶
Back to Proposition III.15.
It will be observed that Euclid’s proof differs from that given in our text-books (which is Simson’s) in that Euclid introduces another line MN, which is drawn so as to be equal to BC but at right angles to EK and therefore parallel to FG. Simson dispenses with MN and bases his proof on a similar proof by Theodosius ($Sphaerica$ I. 6). He proves that the sum of the squares on EH, HB is equal to the sum of the squares on EK, KF; whence he infers that, since the square on EH is less than the square on EK, the square on BH is greater than the square on FK. It may be that Euclid would have regarded this as too complicated an inference to make without explanation or without an increase in the number of his axioms. But, on the other hand, Euclid himself assumes that the angle subtended at the centre by MN is greater than the angle subtended by FG, or, in other words, that M, N both fall outside the triangle FEG. This is a similar assumption to that made in III. 7, 8, as already noticed; and its truth is obvious because EM, EN, being radii of the circle, are greater than the distances from E to the points in which MN cuts EF, EG, and therefore the latter points are nearer than M, N are to L, the foot of the perpendicular from E to MN.
Simson adds the converse of the proposition, proving it in the same way as he proves the proposition itself.