There's this one proof that's been bugging me and I can't seem to get it at all. Given: Isosceles triangle ABC (A being the vertex) and line AF as the < bisector of <BAC's exterior angle. Prove: Line AF is parallel to base BC I have no clue where to start on this...I tried making two congruent triangles but don't have enough information to prove they are congruent so I don't know what to do now. I'd really appreciate some help. Thanks.