If a trisector of an exterior angle of a triangle is parallel to a bisector of an interior angle prove that the other trisector of the exterior angle is parallel to the trisector of an interior angle.
Let the triangle be ABC with exterior angle EAF trisected by FA and GA. Angle ABC is bisected by BD. It is given that $\ AF||BD$.
$ \angle EAF= \angle ABD (vert.\ opp.\ angles)$
$ \frac {\angle B+\angle C}{3}= \frac{\angle B}{2} $
Sovling this we get $\angle B= 2\angle C$.
I don't know how to proceed further. Can someone help me with this question.
As you can see from the diagram below, trisector $AG$ of an external angle of triangle $ABC$ is parallel to bisector $BD$. The other trisector $AF$ is then parallel to $BC$, not to a trisector of an interior angle, because from ${1\over3}(\angle B+\angle C)={1\over2}\angle B$ it follows $\angle C={1\over2}\angle B=\angle EAF$.
The theorem is then, in general, false.