# Stewart's Theorem

## Statement

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If a cevian of length t is drawn and divides side a into segments m and n, then

$c^{2}n + b^{2}m = (m+n)(t^{2} + mn)$

## Proof

For this proof we will use the law of cosines and the identity $\cos{\theta} = -\cos{180 - \theta}$.

Label the triangle $ABC$ with a cevian extending from $A$ onto $BC$, label that point $D$. Let CA = n Let DB = m. Let AD = t. We can write two equations:

• $n^{2} + t^{2} - nt\cos{\angle CDA} = b^{2}$
• $m^{2} + t^{2} + mt\cos{\angle CDA} = c^{2}$

When we write everything in terms of $\cos{\angle CDA}$ we have:

• $\frac{n^2 + t^2 - b^2}{nt} = \cos{\angle CDA}$
• $\frac{c^2 - m^2 -t^2}{mt} = \cos{\angle CDA}$

Now we set the two equal and arrive at Stewart's theorem: $c^{2}n + b^{2}m=m^{2}n +n^{2}m + t^{2}m + t^{2}n$