# Thread: Not sure how to approach.

1. ## Not sure how to approach.

In the figure AD is the external bisector of angle EAC intersects BC produces in D. If AB=8, AC=6, BC=3, find CD.

2. ## Re: Not sure how to approach.

Originally Posted by hisajesh
In the figure AD is the external bisector of angle EAC intersects BC produces in D. If AB=8, AC=6, BC=3, find CD.
Well, you could "brute force" it.

You can find angles BAC and BCA since you know the sides of triangle ABC. From angle BAC you can find its supplement CAE. And CAE is split into two equal parts: CAE and EAD. So you have angle CAD. You can also find the supplement to angle BCA. Then you have two angles and a side of triangle ACD. Now go from there.

It sounds worse than it is....

-Dan

3. ## Re: Not sure how to approach.

You can also use the external angle bisector theorem.

4. ## Re: Not sure how to approach.

I think external angle bisector theorem is a much better option. It just would give
AB/AC = BD/CD and now we can plug in values and get the result.

5. ## Re: Not sure how to approach.

Originally Posted by ibdutt
I think external angle bisector theorem is a much better option. It just would give
AB/AC = BD/CD and now we can plug in values and get the result.

cosine law used twice defines angles of ABC