354 
HERON OF ALEXANDRIA 
reversing the raj^ so that the eye is at D instead of G, and the 
object at C instead of D, we must have (3 > a. But (3 was 
less than a, which is impossible. (Similarly it can be proved 
that a is not less than (3.) Therefore a = /?. 
In the Pseudo-Euclidean Catojotrica 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 A'B). It follows instantaneously (Prop. 1) 
that Z CAE = Z DAH. 
If the mirror is the convex side of a circle, the same result 
follows a fortiori. Let CA, AD meet 
the arc at equal angles, and CB, BD at 
unequal angles. Let AE be the tan 
gent at A, and complete the figure. 
Then, says Heron, (the angles GAG, 
BAD being by hypothesis equal), if we 
subtract the equal angles GAE, BAF 
from the equal angles GAG, BAD (both 
pairs of angles being ‘ mixed be it 
observed), we have Z EAC — Z FA I). Therefore CA + AD 
< GF + FD and a fortiori < GB + 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 yolytheoron 
(‘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’).