Let $AM = x\text{ cm}$. Consider $\Delta ABC$, we have
$\begin{array}{rcl}
AC & = & \sqrt{8^2 + 6^2} \\
& = & 10 \text{ cm}
\end{array}$
Then, $MC = (10-x)\text{ cm}$.
Consider $\Delta AMD$, we have
$\begin{array}{rcl}
AD^2 & = & AM^2 + DM^2 \\
DM^2 & = & 6^2 – x^2 \\
& = & (36-x^2) \text{ cm}^2 \ldots \unicode{x2460}
\end{array}$
Consider $\Delta CMD$, we have
$\begin{array}{rcl}
CD^2 & = & DM^2 + CM^2 \\
DM^2 & = & 8^2 – (10-x)^2 \\
& = & 64 – 100 + 20x -x^2 \\
& = & (-x^2 + 20x – 36) \text{ cm}^2 \ldots \unicode{x2461}
\end{array}$
By solving $\unicode{x2460}$ and $\unicode{x2461}$, we have
$\begin{array}{rcl}
36-x^2 & = & -x^2 + 20x – 36 \\
20x & = & 72 \\
x & = & \dfrac{18}{5}
\end{array}$
Therefore, we have
$\begin{array}{rcl}
AM : MC & = & \dfrac{18}{5} : 10 – \dfrac{18}{5} \\
& = & \dfrac{18}{5} : \dfrac{32}{5} \\
& = & 9 : 16
\end{array}$
