(i) This is an immediate consequence of Casey's theorem. We provide an elementary proof as follows assuming the result in part (ii) (which will be proved later).
Let $a = BC$, $b = CA$ and $c = AB$. By part (ii), we have $BM = BC = a$. Then we have
$$\begin{aligned} AN &= AM = c - a, \\ CN &= b - AN = a + b - c, \\ DM &= AD - AM = \frac{b^2}{c} - c + a. \end{aligned}$$
It follows that
$$\begin{aligned} &BD \cdot CN + BC \cdot DM - CD \cdot BM \\ &= \frac{a^2}{c}(a+b-c) + a\left(\frac{b^2}{c} - c + a\right) - \frac{ab}{c} \cdot a \\ &= \frac{a}{c}(a^2 + ab - ac + b^2 - c^2 + ac - ab) = 0 \end{aligned}$$
using Pythagoras' theorem. This proves the equality.

(ii) Let $O$ and $O'$ be the centres of $\omega$ and $\omega_1$ respectively. Note that $BC$ is a diameter of $\omega$ since $\angle BDC = 90^\circ$. Therefore, the tangent at $B$ to $\omega$ is parallel to the tangent at $N$ to $\omega_1$. By homothety, we know that $B, T, N$ are collinear. As $\angle BTC = 90^\circ$, we have $\triangle BTC \sim \triangle BCN$. Therefore,
$$BC^2 = BT \times BN = BM^2.$$
This gives $BC = BM$.