exterior angles of triangle

The exterior angle of an angle of triangle is greater than both otherangles 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 $\wedge ACD>\wedge ABC$.   $\mathrm{\square }$