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  $\wedge ACD>\wedge ABC$.   $\mathrm{\square }$

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