Stewart’s Theorem

Brute force is an ugly but indispensable method in the math olympian’s toolbox. In the context of Euclidean geometry, brute force is better known as “side/angle/trigo-whacking”, depending on what sorts of quantities are being calculated. For example, you could label the 3 sides of a triangle as a, b and c, then proceed to express every other possible side length or angle in terms of those 3 variables.

One useful theorem that can help the “whacking” process is Stewart’s Theorem, which allows us to express the length of a cevian (i.e. a line segment joining a vertex of the triangle to a point on the opposite side of the triangle). Stewart’s Theorem is easy to state:

Stewart Theorem

Stewart’s Theorem. In the diagram above, the length of the cevian AD is given by the formula

b^2n + c^2m = a(d^2 + mn).

I can never remember the equation above. Instead, I remember the proof in one line and derive the theorem!

Proof: “Cosine rule for the angles at point D.”

Note that \cos ADB = -\cos ADC. By the cosine rule for \triangle ADB, we have:

\begin{aligned} c^2 &= d^2 + n^2 - 2dn \cos ADB, \\  \cos ADB &= \displaystyle\frac{d^2 + n^2 - c^2}{2dn}. \end{aligned}

By the cosine rule for \triangle ADC, we have:

\begin{aligned} b^2 &= d^2 + m^2 - 2dm \cos ADC, \\  \cos ADC &= \displaystyle\frac{d^2 + m^2 - b^2}{2dm}. \end{aligned}

Hence, we have the following:

\begin{aligned} \cos ADB &= - \cos ADC, \\  \displaystyle\frac{d^2 + n^2 - c^2}{2dn} &= \frac{b^2 - d^2 - m^2}{2dm}, \\  m(d^2 + n^2 - c^2) &= n(b^2 - d^2 - m^2), \\  b^2n + c^2m &= m(d^2 + n^2) +n(d^2 + m^2) \\  &= d^2(m+n) + mn(n+m) \\  &= (m+n)(d^2 + mn) \\  &= a(d^2 + mn).\end{aligned}

Done!

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One Response to Stewart’s Theorem

  1. Joseph Nebus says:

    Cevian is a new term to me. Thank you for it.

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