# 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  $\wedge ACD\;>\;\wedge ABC$.   $\Box$

## References

• 1 Karl Ariva: Lobatsevski geomeetria.  Kirjastus “Valgus”, Tallinn (1992).
Title exterior angles of triangle ExteriorAnglesOfTriangle 2013-05-05 8:44:43 2013-05-05 8:44:43 pahio (2872) pahio (2872) 2 pahio (2872) Theorem msc 51M05