jmc

Let's draw a sketch first. Since $△ABC$ is isosceles, we know that $AM$ must form a right angle with $BC.$ We know that $BM=MC=\frac{BC}{2}=5.$ Now we just simply apply the Pythagorean Theorem on the right triangle $△ABM.$ $$\begin{array}{rl}A{M}^2& =A{B}^2-B{M}^2\\ A{M}^2& =8^2-5^2=39\\ AM& =\boxed{\sqrt{39}}\end{array}$$