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May 17, 2014

Forms of Stewart’s theorem

Filed under: mathematics — ckrao @ 10:41 am

Stewart’s theorem finds the length of a cevian d=AD in terms of the side lengths of the triangle ABC and the lengths m, n into which point D on BC divides that side. Here are some forms of the same formula.

Stewart

1. The most common form we see is

\displaystyle b^2 m + c^2 n = a(d^2 + mn) \quad \Rightarrow \quad d^2 = \frac{b^2m + c^2n}{a} - mn.\quad\quad(1)

An easy-to remember form of this is rewriting the above as man + dad = bmb + cnc (a man and his dad hid a bomb in the sink!). This can be proved by applying the cosine rule to triangles ACD and then ABC:

\begin{aligned}  d^2 &= AD^2\\  &= AC^2 + CD^2 - 2AC.CD\cos \angle DCA\\  &= AC^2 + CD^2 - 2AC.CD \cos \angle BCA\\  &= AC^2 + CD^2 - 2AC.CD \frac{CA^2 + CB^2 - AB^2}{2CA.CB}\\  &= b^2 + n^2 - 2bn \frac{b^2 + a^2 - c^2}{2ba}\\  &= b^2 + n^2 - n \frac{b^2 + a^2 - c^2}{a}\\  &= \frac{b^2(m+n) - n(b^2 + a^2 - c^2)}{a} + n^2\\  &= \frac{b^2 m + c^2n}{a} + n^2 - \frac{na^2}{a}\\  &= \frac{b^2 m + c^2n}{a} + n(n-a)\\  &= \frac{b^2 m + c^2n}{a} - mn.\\  \end{aligned}

2. If D divides the side BC in the ratio BD:DC = r:s,

\displaystyle d^2 = \frac{rb^2 + sc^2}{r+s} - \frac{rsa^2}{(r+s)^2} = \frac{r^2b^2 + s^2c^2 + rs(b^2 + c^2 - a^2)}{(r+s)^2} .\quad\quad(2)

3. Similar to (2) but substituting \displaystyle \lambda = r/(r+s) = BD:BC,

\displaystyle d^2 = \lambda b^2 + (1-\lambda)c^2 - \lambda(1-\lambda)a^2.\quad\quad(3)

This and the previous form are conveniently proved using vectors. Writing the vector \mathbf{AD} = \lambda \mathbf{AC} + (1-\lambda) \mathbf{AB},

\begin{aligned}  d^2 &= \mathbf{AD}. \mathbf{AD}\\  &= \left(\lambda \mathbf{AC} + (1-\lambda) \mathbf{AB}\right).\left(\lambda \mathbf{AC} + (1-\lambda) \mathbf{AB}\right)\\  &= \lambda^2 \mathbf{AC}.\mathbf{AC} + (1-\lambda)^2 \mathbf{AB}.\mathbf{AB} + 2\lambda(1-\lambda)\mathbf{AB}.\mathbf{AC}\\  &= \lambda^2 b^2 + (1-\lambda)^2 c^2 + \lambda(1-\lambda)\left(b^2 + c^2 - a^2\right)\\  &= b^2(\lambda^2 + \lambda(1-\lambda)) + c^2((1-\lambda)^2 + \lambda(1-\lambda)) - a^2\lambda(1-\lambda)\\  &= \lambda b^2 + (1-\lambda)c^2 - \lambda(1-\lambda)a^2.  \end{aligned}

Note that this is valid for any real \lambda, so D may lie beyond segment BC.

4. Writing (3) as a quadratic in \lambda:

\displaystyle d^2 = \lambda^2 a^2 + \lambda(c^2-b^2 -a^2) + b^2.\quad\quad(4)

5. A symmetric form [1], where the following distances are taken as directed segments (CD = -DC etc.)

\displaystyle \frac{BA^2}{BC.BD} + \frac{CA^2}{CB.CD} + \frac{DA^2}{DB.DC} = 1\quad\quad(5)

 Note that this is equivalent to BA^2 . CD + CA^2 . DB = DA^2.CB + CD.DB.CB = CB(DA^2 + CD.DB) which is (1).

 

Here are a few special cases of this formula applying form (3).

  • D = C (i.e. \lambda = 1): d^2 = b^2
  • D is a third of the way along CB (closer to C) (\lambda = 2/3): \displaystyle d^2 = 2b^2/3 + c^2/3 - 2a^2/9
  • AD is the internal angle bisector of \angle BAC (\lambda = c/(b+c)):

\begin{aligned}  d^2 &= cb^2/(b+c) + bc^2/(b+c) - bca^2/(b+c)^2\\  &= bc\left[\frac{b}{b+c} + \frac{c}{b+c}- \left(\frac{a}{b+c}\right)^2\right]\\  &= bc\left[1 - \left(\frac{a}{b+c}\right)^2\right]\\  &= bc - mn.  \end{aligned}

  • AD is the external angle bisector of \angle BAC (assume b > c so \lambda = -c/(b-c)):

Stewart external bisector

\begin{aligned}  d^2 &= \frac{-c}{b-c} b^2 + \frac{b}{b-c}c^2 + \frac{bc}{(b-c)^2}a^2\\  &= bc\left[\frac{-b}{b-c} + \frac{c}{b-c}+ \left(\frac{a}{b-c}\right)^2\right]\\  &= bc\left[ \left(\frac{a}{b-c}\right)^2 - 1\right]\\  &= mn - bc.  \end{aligned}

Reference

[1] Darij Grinberg – Unpublished Notes – Geometry

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