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

  • Thread starter Scorpino
  • Start date
In summary, the conversation is about proving that line AF is parallel to base BC in an isosceles triangle, using the Isosceles triangle theorem and angle addition. The individual is asking for help and someone explains that the two new angles created by line AF are the angle bisectors. They then go on to show how to isolate the angles and prove that AF and BC are parallel.
  • #1
Scorpino
7
0
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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  • #2
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.
 
  • #3
wait what are these two new angles that are created with line AF?

EDIT: Nevermind I understand...they're the angle bisectors. Thanks!
 
  • #4
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?
 

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

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