Solution:

Answer: $\frac{25}{9}$. Quadrilaterals $AMID$ and $DICT$ are cyclic, having right angles $\angle AMD$, $\angle AID$, and $\angle CID$, $\angle CTD$ respectively. We see that $M$, $I$, and $T$ are collinear. For, $m \angle MID = \pi - m \angle DAM = \pi - m \angle DAB = m \angle BCD = \pi - m \angle DCT = \pi - m \angle DIT$. Therefore, Menelaus' theorem applied to triangle $MTB$ and line $ICA$ gives
$$\frac{MI}{IT} \cdot \frac{TC}{CB} \cdot \frac{BA}{AM} = 1$$
On the other hand, triangle $ADM$ is similar to triangle $CDT$ since $\angle AMD \cong \angle CTD$ and $\angle DAM \cong \angle DCT$ and thus $AM / CT = AD / CD$. It follows that
$$\frac{MI}{IT} = \frac{BC \cdot AM}{AB \cdot CT} = \frac{BC \cdot AD}{AB \cdot CD} = \frac{5 \cdot 10}{3 \cdot 6} = \frac{25}{9}$$