By Euclid's exterior angle theorem, any exterior angle of a triangle is greater than either of the interior angles at the opposite vertices:
32.
There are some elongated black costal points and a black streak by the interior angle and another near the base of the interior border.
33.
There is a white bent transverse line extending from the fourth mark to the interior angle and the marginal points are black and minute.
34.
Now the interior angle of the triangle " OPR " at O is one third of the triangle's exterior angle at " R ".
35.
Then one of the alternate angles is an exterior angle equal to the other angle which is an opposite interior angle in the triangle.
36.
To prove proposition 29 assuming Playfair's axiom, let a transversal cross two parallel lines and suppose that the alternate interior angles are not equal.
37.
As the proof only requires the use of Proposition 27 ( the Alternate Interior Angle Theorem ), it is a valid construction in absolute geometry.
38.
This implies that there are interior angles on the same side of the transversal which are less than two right angles, contradicting the fifth postulate.
39.
The interior angle concept can be extended in a consistent way to crossed polygons such as star polygons by using the concept of directed angles.
40.
Therefore, it would make sense for a 180 degree angle to have an interior angle of 180 degrees, and an exterior angle of 0 degrees.
