In $\triangle A B C, B D$ is the median
Given that, $a=5, b=6, c=7$
We know that,
$$
\begin{array}{rlrl}
B D^2=\frac{1}{4}\left[2 c^2\right. & \left.+2 a^2-b^2\right]=\frac{1}{4}\left[2(7)^2+2(5)^2-6^2\right] \\
& = & \frac{1}{4}[98+50-36]=\frac{1}{4}[98+14] \\
\Rightarrow & B D^2 & =28 \\
\Rightarrow & & B D & =2 \sqrt{7}
\end{array}
$$