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Prove that the external bisector of an angle of a triangle (not isosceles) divides the opposite side (externally) into two segments proportional to the sides of the triangle adjacent to the angle enter image description here

That is, show $\dfrac{BA}{BC} = \dfrac{B'A}{B'C}$

Here are my questions regarding the wording of this question.

  1. So the triangle (not isosceles) drawn is the black one $\triangle ABC$. The external bisector is the one to angle A. Now when it says "divides the opposite side (externally)", what do they mean by "opposite" here? How do I know that they mean side $BC$ is the opposite side they are referring to? Answered by Andres

  2. I am also guessing when they say two segments, they aren't referring to the trivial side $BC$?

  3. Also when they say "two segments proportional to the sides of the triangle adjacent to the angle", what do they mean by "adjacent to the angle"? What does adjacent here mean? And why is it $AC$ and $AB$? My thinking is that they mean the side that is beside the angle that's being bisected? That is, side $AC$ is one of them? But how is $AB$ side the x subtended by the blue line?

  4. When is "(externally)" in parenthesizes? In Geometry are "extended sides" also considered as "sides"?

If I were to be explicit, is this also equivalent?

$$B'C = \alpha BA$$

$$B'A = \alpha BC$$

Dividing them out gives me the result. But how can i assume that the proportionality constant is the same from the wording of the problem

EDIT After a talk with my professor, he just said it's a language and convention accepted. He couldn't give me a better explanation other than that one. So problem is "solved"

Please don't solve the problem. I just want to get used to reading these questions

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The picture could be better. The blue line should be a continuation of $BA$. The "opposite" side is, as you conjecture, $BC$, since the other two sides are met by the external bisector precisely at $A$. –  André Nicolas Oct 19 '12 at 5:52
    
@AndréNicolas, you are correct. I'll update my question because I've managed to clarify some parts –  Hawk Oct 19 '12 at 20:07
    
OK. I expect you are familiar with the analogue for internal angle bisector. –  André Nicolas Oct 19 '12 at 22:23

2 Answers 2

up vote 0 down vote accepted

Where did you get that problem? When the angles at B and C are 45 and 110 degrees respectively, I got in a drawing AB=12.7, AC=9.6, BE=24.5 and BC=5.8 and the relation you gave is evidently not satisfied.

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mathworld.wolfram.com/ExteriorAngleBisector.html for picture –  Hawk Oct 19 '12 at 20:05
    
But you have edited you post and changed your original relation AB/BC=BE/EC. –  Mok-Kong Shen Oct 20 '12 at 10:49
    
It's the same picture and I just changed the bisector names and points of intersection –  Hawk Oct 20 '12 at 18:19
    
No, you not only changed the naming of the points but also the equation. I was referring to your original post which doesn't correspond to the current one in content. But it isn't worthwhile to argue about things that no longer exist to be verified. –  Mok-Kong Shen Oct 22 '12 at 13:09

Make line parallel to B'B from A and meet at A' on BC. Angle BAA'= BA'A = x. So BA=BA'. Since parallel divide B'C and BC as B'C:B'A = BC:BA' = BC:BA

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