Stewart's theorem: If a cevian AD of length d divides side BC into segments m (BD) and n (DC), then $b^2$*m + $c^2$n - $d^2$a = amn. For a median (m = n = a/2): the median length formula is $m_a$ = $\frac{1}{2}$*sqrt(2$b^2$ + 2$c^2$ - $a^2$). This is occasionally tested in JEE.