China Hong Kong Mathematical Olympiad

(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 $BC = BM$.