THE CATOPTRIC A
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cylindrical mirrors play a part in these arrangements. The
whole theory of course ultimately depends on the main pro
positions 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 1 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.
N ow Z FAF = Z BA D
= Z CAE, by hypothesis.
Therefore the triangles AEF, A EC, having two angles equal
and AE common, are equal in all respects.
Therefore G A = AF, and GA + AD = DF.
Since FE = EC, and BE is perpendicular to FC, BF = BG.
Therefore CB + BD = FB + BD
> FD,
i.e. > CA +AD.
The proposition was of course known to Archimedes. We
gather from a scholium to the Pseudo-Euclidean Catoptrica
that he proved it in a different way, namely by reductio ad
absurdum, thus : Denote the angles CAE, DAB by a, (3 re
spectively. Then, a is > = or < (3. Suppose a > /3. Then,
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