# 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!

## References

- 1 Karl Ariva: Lobatsevski geomeetria. Kirjastus “Valgus”, Tallinn (1992).

Title | sum of angles of triangle in Euclidean geometry |
---|---|

Canonical name | SumOfAnglesOfTriangleInEuclideanGeometry |

Date of creation | 2013-09-26 10:00:16 |

Last modified on | 2013-09-26 10:00:16 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 4 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 51M05 |