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Consider the triangle defined by $ A(3, -5)$, $B(1, -3) $ and $ C(2, -2) $. I need to get the length of the angle B external bisector. Here my picture enter image description here

I think I know the process:

  • Find the distances
  • Get the ratio using the exterior bisector theorem
  • Use the ratio formula to get abscissa and ordinate
  • And finally get the BD distance

The distances:

$ AB = {\sqrt 8},\quad BC = {\sqrt 2} $

The ratio:

$ \frac{AB}{BC} = \frac{DA}{DC},\quad \frac{\sqrt 8}{\sqrt 2} = \frac{DA}{DC}\quad \frac{DA}{DC} = 2 $

But I'm struggling with the ratio. If I change $ \frac{AB}{BC} = \frac{DA}{DC} $ to $ \frac{AB}{BC} = \frac{AD}{CD} $, I will get the exact same ratio, but different differences in the ratio formula: $$ r = \frac {x - x_1}{x_2 - x}$$

Ok, that's the expected thing. But the question is:

What is the correct order for the bisector theorem when using it with ordered pairs?

Maybe from left to right, or from the bigger side to the short one? I don't know.

PD: I would like to stick with this approach since I'm still practicing with the very basics, as you can see :D And the main question is the order for the bisector theorem, not the bisector distance actually.

Jose Carlos Ramírez Vásquez
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1 Answers1

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$2 = \frac {x-3}{x-2}\\ 2x - 4 = x- 3\\ x = 1$

enter image description here

Doug M
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  • Thanks for noticing that! I made a test to check if it always end with the same result no matter the order, the result was it returns the same value only if the order is changed in both the numerator AND the denominator: http://imgur.com/a/KfLmF – Jose Carlos Ramírez Vásquez Jun 27 '17 at 00:04
  • But going back to the question problem, when I use $ \frac {DA}{DC} $ and $ \frac {AD}{CD} $ the result is $ 1 $ but when I use $ \frac {AD}{DC} $ and $ \frac {DA}{CD} $ the result is $ 7/3 $ ... so the question again? which one is the direction should be used? – Jose Carlos Ramírez Vásquez Jun 27 '17 at 00:12
  • I suggest that you do not try to memorize formulas for geometry problems. Always construct the figure. Then you can see what the result is that you are looking for. This is a pretty obscure theorem, and not worth the effort. You are measuring from $D$ to $C$ and from $D$ to $A.$ The ratio $\frac {AD}{DC}$ would place $D$ between $A$ and $C.$ – Doug M Jun 27 '17 at 00:25
  • I didn't notice about that either! Thanks a lot, I think I should keep practicing, haha, thanks for the suggestion about always construct the figure :-) anything else would you advice regarding this topic? – Jose Carlos Ramírez Vásquez Jun 27 '17 at 00:40