Full text: From Aristarchus to Diophantus (Volume 2)

THE CATOPTRIC A 
353 
« f 
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, 
1623.2 A a
	        
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