# Not sure how to approach.

• Sep 27th 2013, 03:03 PM
hisajesh
Not sure how to approach.
In the figure Attachment 29309 AD is the external bisector of angle EAC intersects BC produces in D. If AB=8, AC=6, BC=3, find CD.
• Sep 27th 2013, 03:19 PM
topsquark
Re: Not sure how to approach.
Quote:

Originally Posted by hisajesh
In the figure Attachment 29309 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
• Sep 27th 2013, 04:20 PM
emakarov
Re: Not sure how to approach.
You can also use the external angle bisector theorem.
• Sep 27th 2013, 10:37 PM
ibdutt
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.
• Sep 28th 2013, 07:00 AM
bjhopper
Re: Not sure how to approach.
Quote:

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