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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.

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  • $\begingroup$ It is not clear where E is located. Is it a point on BA? $\endgroup$ – Mick Sep 3 '14 at 2:26
  • $\begingroup$ I had this problem too. The text is vague, but I am supposing it is on (segment)AB. $\endgroup$ – Chris Swanson Sep 3 '14 at 3:07
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For some reason, ACE is an isosceles triangle with AC = AE.

It is easier to prove BD/CD = BD'/CD' instead.

By angle bisector theorem, BD/CD = AB/AC

(See above), AB/AC = AB/AE

By basic proportional theorem, AB/AE = BD'/ CD'

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  • $\begingroup$ This is great, but how do we know ACE is isosceles? In fact, for this to be the case then (segment)AD is perpendicular to (line)CE. $\endgroup$ – Chris Swanson Sep 3 '14 at 3:06
  • $\begingroup$ You are correct! If we set (segment)AD as a transversal to (line)AD' and (line)CE then we may use the Alternate Interior of Parallel lines theorem. The bisector segments(AD) and (segment)AD' form four angles of which two pairs are both congruent. What we see from the interior angles is they are the sum of half of one bisected angle and half of the other bisected angle. Hence, two angles of the transversal and one parallel line form a linear pair and are congruent. Hence, they are right angles. BRILLIANT! $\endgroup$ – Chris Swanson Sep 3 '14 at 3:13
  • $\begingroup$ @ChrisSwanson Right. You got it. In fact, the angle between the interior and exterior angle bisector of an angle is always 90 degrees. $\endgroup$ – Mick Sep 3 '14 at 14:05
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I completed the problem using the gracious information from Mick so if anybody who is looking for a similar answer that ins't completely aware of the context of the answers they may follow this proof.

If there are any errors please let me know so I may correct them.

(Please excuse my poor latex skills.)

Big Thanks to Mick.

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