A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

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
354 HERON OF ALEXANDRIA
reversing the ray so that the eye is at D instead of 0, and the
object at C instead of D, we must have fi > a. But (3 was
less than a, which is impossible. (Similarly it can be proved
that a is not less than ft.) Therefore a. = /?.
In the Pseudo-Euclidean Gatoptrica the proposition is
practically assumed ; for the third assumption or postulate
at the beginning states in effect that, in the above figure, if A
be the point of incidence, CE : EA = DH : HA (where DH is
perpendicular to AB). It follows instantaneously (Prop. 1)
that
ACAE = LDAH.
If the mirror is
the convex side of a circle, the same result
follows a fortiori. Let GA, AD meet
the arc at
equal angles, and CB, BD at
unequal angles. Let AE be the tangent
at A, and complete the figure.
Then, says Heron, (the angles GAC,
BAD being by hypothesis equal), if
we
subtract the equal angles GAE, BAF
from the equal angles GAC, BAD (both
pairs of angles being ' mixed
', be it
observed), we have Z EAC = I FAD. Therefore CA+AD
< CF+FD and a fortiori < CB + BD.
The problems solved (though the text is so corrupt in places
that little can be made of it) were such as the following:
11, To construct a right-handed mirror (i.e. a mirror which
makes the right side right and the left side left instead of
the opposite); 12, to construct the mirror called polyiheoron
('with many images'); 16, to construct a mirror inside the
window of a house, so that you can see in it (while inside
the room) everything that passes in the street;
18, to arrange
mirrors in a given place so that a person who approaches
cannot actually see either himself or any one else but can see
any image desired (a 'ghost-seer').