An angle bisector of a triangle divides the opposite side into..
I had a two part problem. I did (a) with no difficulty, but I'm stuck on (b) especially with how we are asked to prove it.
(a) Prove that the external angle bisector at C is parallel to line AB if and only if AC = BC. No problem, I did this.
(b) If AC is not equal to BC, the external angle bisector at C intersects line AB at a point Q (by part a) Use the Law of Sines to prove that QA/QB = AC/BC.
Here is a link for the same problem, but with different choice of letters. The part of this link that the problem is on is SupThm8 where it states if triangle ABC is not isosceles. There have been several hints saying to draw parallel lines, but how would I do this with the law of sines!?
Re: An angle bisector of a triangle divides the opposite side into..
A GOOD question. Attachment 26876The solution is attached herewith.