166 On the Theory of 
Prop. II. Fig. 5. 
The three angles of a triangle cannot be greater than two 
right angles. 
If it be possible, let the three angles of the triangle ABC be 
greater than two right angles, and let the excess above two right 
angles be equal to the angle x. Construct the triangle BCF, 
having the sum of its angles equal to the sum of tlie angles of 
the triangle ABC, and one angle F BC equal to, or less than, 
half the angle ABC; in like manner, construct another triangle 
F’B’C’ having the sum of its angles equal to the sum of the an- 
gles of the triangle F BC, and one angle F’B’C’ equal to, or less 
than, half the angle F BC ; and continue the like constructions 
as far as necessary. Because the angle F B C is equal to, or less 
than, half ABC ; and the angle F’B’C’, equai to, or less than, 
half F BC, and so on; by continuing the series of triangles far 
enough, we shall at length arrive at one, viz. F’B’C’, having an 
angle F’B’C’ less than the given angle x*. And because the 
three angles of every triangle in the series make the same sum, 
the three angles B’C’F’, B’F'C’, F’B’C’ will be together equal to 
the sum of two right angles and the angle x: but the angle 
F’B/C’ is less than the angie 2; wherefore the angles B’C’F’ 
and B’F’C’ are greater than two right angles; which is absurd 
(17.1.E.) Therefore the three angles of a triangle cannot ba 
greater than two right angles. 
The following demonstration does not fall off from the accu- 
racy and spirit of the ancient geometry, although, for the sake 
of brevity, it is not dressed out in the usual costwme. 
Prop. III. Fig. 6. 
The three angles of any triangle are equal to two right angles. 
If what is affirmed be not true, let the three angles of the tri- 
angle ACB be less than two right angles, and let the defect from 
two right angles be equal to the angle a. Let P stand fer a 
right angle, and find a multiple of the angle x, viz. m x x, such 
that 4P—m x a, or the excess of four right angles above the 
multiple angle, shall be less than the sum of the two angles AC B 
and ABC of the proposed triangle. Produce the side C B, and 
cut off BE, EF, FG, &c. each equal to BC, so that the whole 
CG shali contain CB m times; and construct the triangles 
BHE, EKF, FLG, &c., having their sides equal to the sides 
of the triangle AC B, and consequently, their angles equal to 
the angles of the same triangle. In CA produced, take any 
point M, and draw HM, KM, LM, &c.; AH, HK, KL, &c. 
All the angles of all the triangles into which the quadrilateral 
* For by continually bisecting any proposed magnitude, a magnitude 
will at length be found less than any given maguitude. 
figure 
