Jan
29
2 minute read

I just ran into a few very simple extensions of the angle bisector theorem which I’ve never noticed before. Since I couldn’t easily find this anywhere online it seems interesting enough to note here.

Subsegment Lengths

If DB is the angle bisector of triangle ABC and side AC = b, side BC = a and side AB = c

Then   $ m= \dfrac{a \cdot b}{a+c} $ and $ n = \dfrac{b \cdot c}{a+c} $

This is proved as following:

  • From the angle bisector Theorem $ \dfrac{m}{n} = \dfrac{a}{c} $
  • So $ m = n \cdot \dfrac{a}{c} $
  • Then substitute in n = b - m   to get $ m = (b - m) \dfrac{a}{c} $

  • Solve for m:    $ (1 + \dfrac{a}{c}) \cdot m = \dfrac{ab}{c} $

  • That then simplifies to $ m = \dfrac{a \cdot b}{a+c} $ 

Bisector Length

A second also useful extension is that we can derive an expression for the bisector BD in terms of  other segments

  • Let  side AC =  a,   BC = a, BD = d, AC = b AD = b and CD = n as labelled above.
  • Using Stewart’s Theorem we get $ a^2 \cdot m + c^2 \cdot n = b (m \cdot n + d^2) $ but this can simplified.
  • From the angle bisector theorem $ a = \dfrac{cn}{m} $ and $ c  = \dfrac{am}{n} $
  • Substituting those into the left hand side we get:  $ acn + acm = b (m \cdot n + d^2) $
  • Combine:   $ acn + acm  = (m + n)ac = abc $
  • So $ abc = b (m \cdot n + d^2) $ We can then eliminate a b on each side and rearrange:
\[d^2 = a \cdot c  -  m \cdot n\]

Exterior Angle Bisector Theorem

It turns out there is also an exterior angle bisector theorem!  

  • Given a triangle ABC, let angle BAC = x and angle ABC = y.  Then bisect the exterior angle of B as show above and project down to point E.

  • Then create a parallel line AF to BC.

  • By angle chasing : $ \angle{CBE} = \dfrac{180 -y} {2} = 90 - \dfrac{y}{2}$ 
  • $ \angle{BCE} = x + y $
  • $ \angle{BAF} = \angle{FAC} - \angle{BAC}  = x + y  - x = y $
  • $\angle{ABF} = 180 - (\angle{ABC} + \angle{CBE}) =  180 - (y +   90 - \dfrac{y}{2}) =   90 - \dfrac{y}{2}  $
  • $ \angle{AFE} = 180 - (\angle{ABF} + \angle{BAF}) = 180 - (  90 - \dfrac{y}{2}  + y) =   90  - \dfrac{y}{2} $

So Triangle ABF turns out to be isosceles and AF = AB = c.  Since triangle AFE is similar to CBE (from our original parallel line) we finally get the desired ratio:

$\dfrac{d}{a} = \dfrac{b+d}{c}$

Exterior Bisector Lengths

Its probably not too surprising that we can apply Stewart’s law again and get a formula for all the lengths above:

We start with:  $e^2 \cdot b + c^2 \cdot d = (b + d)(b\cdot d + a^2) $

  • From the ratio above we have $ d = \dfrac{a(b+d)}{c} $

  • Substituting that gives us: $e^2 \cdot b + ac(b+d) = (b + d)(b\cdot d + a^2) $

  • Collecting the (b+d) terms : $e^2  = \dfrac{(b + d)}{b}(b\cdot d + a^2 -  ac) $

  • Note $(b + d)(a - c) =  ab + ad - cd  - bc$ and again from the ratio ab + ad = cd so combined we get  $(b + d)(a - c) =  -bc  $

  • So $e^2  =  (b +d)d  + \frac{a}{b}(b+d)(a-c)  =  (b +d)d   - ac  $  which is very similar to the equivalent angle bisector formula.

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