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 $A B C$ for example the exterior angle $\wedge ACD$ where $D$ is point on the lengthening of the side $B C$ nearer to $C$ than to $B$ . Let $E$ be the midpoint of $A C$ . Let $B E$ be the median of the triangle. We find on its lengthening the point $F$ such that $EF=EB$ . Then the triangles $A B E$ and $C E F$ 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
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