How Can Line AF Be Proven Parallel to Base BC in an Isosceles Triangle?

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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.
 
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Well because of the Isosceles triangle theorem B and C are congruent. Well also know that the new two angles you created with line with line AF will also be congruent. Since the 2 angles and angle A add up to 180 as well as all the angles in the triangle you can see that angle B will be congruent to the Angle FAX (Where X is a point outside A on line AB). Therefore BC and AF are parallel because corresponding angles are congruent.
 
wait what are these two new angles that are created with line AF?

EDIT: Nevermind I understand...they're the angle bisectors. Thanks!
 
Hold on...

4) m<FAX + m<CAF + m<BAC = 180 4) Angle addition
m<CBA + m<ACB + m<BAC = 180

5) m<FAX + m<CAF = m<CBA + m<ACB 5) substitution (after subtracting m<BAC)

How do I isolate FAX and <CBA?