Stewart's Theorem

Consider the diagram.

Suppose that in triangle $ABC$, we draw $D$ on $BC$ and $BD=m$, $AB=c$, $AC=b$, $CD=n$, $m+n=BC=a$, $AD=d$.

Since $ADC = 180- ABD$, we have $\cos ADC = -\cos ABD$.

By the Law of Cosines, $c^2 = d^2 + m^2 - 2dm \cos ADB$, and also $b^2=d^2+n^2 - 2dn \cos ADC$..

We find $\cos ADB = \cos ADC = \frac{d^2+m^2-c^2}{2dm} = -\frac{d^2+n^2-b^2}{2dn}$

(basically we just isolated the cosines in each equation and set them equal)

Now we obtain $d^2n + m^2 n - c^2 n = -d^2m - n^2m + b^2m$ or $d^2n + d^2m + m^2n + n^2m = b^2m + c^2n$, and simplifying yields $d^2a + amn = b^2m + c^2n$, or Stewart's theorem.

For memorization purposes, it may be easier to remember

$$man+dad=bmb+cnc$$

The mnemonic to remember this formula is "A man and his dad put a bomb in the sink."