A history of Greek mathematics - Wilbourhall.org

THE CATOPTRTCA 353
cylindrical mirrors play a part in these arrangements. The
whole theory of course ultimately depends on the main propositions
4 and 5 that the angles of incidence and reflection
are equal whether the mirror is plane or circular.
Herons proof of equality of angles of incidence and reflection.
Let AB be a plane mirror, C the eye, D the object seen.
The argument rests on the fact that nature ' does nothing in
vain'. Thus light travels in a straight line, that is, by the
quickest road. Therefore, even
when the ray is a line broken
at a point by reflection, it must
mark the shortest broken line
of the kind connecting the eye
and the object. Now, says
Heron, I maintain that the
shortest of the broken lines
(broken at the mirror) which
connect G and D is the line, as
CAD, the parts of which make equal angles with the mirror.
Join DA and produce it
to meet in F the perpendicular from
C to AB. Let B be any point on the mirror other than A,
and join FB, BD.
Now
LEAF = L BAD
= Z CAE, by hypothesis.
Therefore the triangles AEF, AEC, having two angles equal
and AE common, are equal in all respects.
Therefore CA = AF, and CA + AD = DF.
Since FE = EG, and BE is perpendicular to FC, BF = BG
Therefore GB + BD = FB + BD
i.e.
> FD,
> GA +AD.
The proposition was of course known to Archimedes.
We
gather from a scholium to the Pseudo- Euclidean Gatoptrica
that he proved it in a different way, namely by reductio ad
absurdum, thus : Denote the angles GAE, DAB by a, /? respectively.
Then, a is > = or < /?. Suppose a > /3. Then,
1623-2 a a