# exterior angles of triangle

The exterior angle^{} of an angle of triangle is greater than both other
angles of the triangle.

Proof. Let us study in an arbitrary triangle $ABC$ for example the exterior angle $\wedge ACD$ where $D$ is point on the lengthening of the side $BC$ nearer to $C$ than to $B$. Let $E$ be the midpoint^{} of $AC$. Let $BE$ be the median of the triangle. We find on its lengthening the point $F$ such that $EF=EB$. Then the triangles $ABE$ and $CEF$ are congruent (SAS). Consequently, we have
$\wedge ECF=\wedge BAE$
and therefore $\wedge ACD>\wedge BAC$. Analogically one shows that
$\wedge ACD>\wedge ABC$. $\mathrm{\square}$

## References

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

Title | exterior angles of triangle |
---|---|

Canonical name | ExteriorAnglesOfTriangle |

Date of creation | 2013-05-05 8:44:43 |

Last modified on | 2013-05-05 8:44:43 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 2 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 51M05 |