Given any triangle ABC, If the bisectors of the interior and exterior angles at A intersect (line)BC at points D and D', respectively, then BD/BD' = CD/CD'.
There is a hint to the problem: introduce (line)CE parallel to (line)AD'. Also, I must use a result from a previous problem: The bisector of an angle of a triangle separates the opposite side into segments whose lengths are proportional to the lengths of the adjacent sides.
These hints yield four proportions:
1) BD/CD = AB/AC from the problem in the hint
2) BC/BD' = BE/BA from the basic proportionality theorem
3) CD'/BD' = AE/AB from the basic proportionality theorem, and
4) BC/CD' = BE/AE from the basic proportionality theorem.
I am having the devil of a time showing the conclusion of the proof which is interesting as one quotient is the familiar result of a partial segment over the entire segment. While the second quotient is a segment over another discrete segment with no union other than a point between the two.
There must be a "link" between the result of problem in the hint and the proportions derived from the basic proportionality theorems but I can not see it.
Any help would be great.