TRIANGLE AND THE SUM OF THE ANGLES. 343 
augles of the triangles DFA and EAF FIG 2 
respectively be S and 8,; S is not D 
greater than S,. For let s be the 
sum of the angles of the triangle ADE ; 
then 
S =F + FAE + EAD+D, 
and, S,= F + FAE + AEF. 
-. $, — § = AEF — (EAD + D),A C rE 
= AEF + AED—(AED+EAD+D), 
=2—s; 
a right angle being taken as the unit of measure. But, by the Pro- 
position, s is not greater than 2. Therefore S is not greater than Sa: 
Cor. 2.—From B, a point within the triangle DAF, draw BC to a 
point C in AF; and let S, be the sum of the angles of the triangle 
ABC. Then 8, is not less than 8. For, produce AB to E; and join 
EC. Then, by Cor. 1, S, is not less than the sum of the angles of 
_ the triangle AEC ; which sum, again, is not less than 8,, or the sum 
of the angles of the triangle AEF; and S, is not less than S. 
Therefore S, is not less than S. 
Proposition II. 
If any triangle CHE (Fig. 3) 
have S, the sum of its angles, FIG 3 
equal to two right angles, every 
triangle has the sum of its an- 
_ gles equal to two right angles. 
For, CE being a side which 
is not less than any other side 
of the triangle CHE, let fall 
HD perpendicular on OE. 
Then HD cannot fall without 
the base CE; else (supposing 2. 
it to fall beyond E) the an- 
gles CEH would be greater than a right angle: hence, because CE 
is not less than CH, the angle CHE would be greater than a right 
angle: so that S would be greater than two right angles: which 
(Prop. I.) is impossible. Produce CD to F; making DF = CD. 
