BOOK I. PROP. XXVI. THEOR.
27
but 
= 
(hyp.)
and therefore 
= which is absurd;
hence neither of the sides 
and 
is greater than the other; and ∴ they are equal;
∴ 
= , and 
= , (pr. 4.).
CASE II.
Again, let = which lie opposite
the equal angles 
and . If it be possible, let
> 
, then take = , draw 
.
Then in 
and 
we have = ,
= and 
= ,
∴ = 
(pr. 4.)
but = (hyp.)
∴ = which is absurd (pr. 16.).
Consequently, neither of the sides or is greater than the other, hence they must be equal. It follows (by pr. 4.) that the triangles are equal in all respects.
Q. E. D.
