sum of angles of triangle in Euclidean geometry

The parallel postulate (in the form given in the parent entry (http://planetmath.org/ParallelPostulate)) allows to prove the important fact about the triangles in the Euclidean geometry:

Theorem.  The sum of the interior angles of any triangle equals the straight angle.

Proof.  Let $ABC$ be an arbitrary triangle with the interior angles $\alpha$, $\beta$, $\gamma$.  In the plane of the triangle we set the lines $AD$ and $AE$ such that  $\wedge BAD=\beta$  and  $\wedge CAE=\gamma$.  Then the lines do not intersect the line $BC$.  In fact, if e.g. $AD$ would intersect $BC$ in a point $P$, then there would exist a triangle $ABP$ where an exterior angle (http://planetmath.org/ExteriorAnglesOfTriangle) of an angle would equal to an interior angle of another angle which is impossible.  Thus $AD$ and $AE$ are both parallel to $BC$.  By the parallel postulate, these lines have to coincide.  This means that the addition of the triangle angles $\alpha$, $\beta$, $\gamma$ gives a straight angle.

See also http://www.cs.bham.ac.uk/research/projects/cogaff/misc/triangle-sum.html#pardoe-proofthis intuitive proof!