646 
Proceedings of the Royal Society 
has the sum of its angles the same, while the sum of A and D = BAC. 
Of these angles one island the other <£ than J A. Taking the least 
of them, and bisecting the opposite side, we derive as before from 
ADC a triangle, still having the same area, and the same sum of all 
the angles, but in which the sum of two of the angles A. 
By a similar process we derive another triangle, still having the 
area and the sum of its angles unaltered, but in which the sum of two 
angles A. 
At last we get a triangle, in which the area is the same as at first, 
and the sum of the angles the same, but the sum of two of them 
the sum of two angles is as small as we please. 
But the third angle can never be greater than 2R, hence the sum 
of the angles of the original triangle cannot be > 2R. 
It is to be noticed that this demonstration would fail if a straight 
line were re-entrant, or if two straight lines had more than one point 
of intersection. 
Corollary . — If C' be the external angle at C of the triangle ABC, 
then, since 
A + B + C = 2R-S, 
where R stands for a right angle, and S is either zero or essentially 
positive, and 
C + C'=2R, 
we have 
C' = A + B + S ; 
That is, the exterior angle of any triangle is not less than the sum 
of the two interior opposite angles. 
Of course it follows that the exterior angle of any triangle is 
greater than either of the interior opposite angles ; and that the sum 
of any two angles of a triangle is less than two right angles. 
We can now prove for hyperbolic space : — 
That the greater side of every triangle has the greater angle oppo- 
site , and conversely. 
That any two sides of a triangle are together greater than the 
third side. 
