## 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’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!