Let $ \triangle ABC$ be a right-angled triangle with $ \angle C=90^\circ$. $ CD$ is the altitude from $ C$ to $ AB$, with $ D$ on $ AB$. $ \omega$ is the circumcircle of $ \triangle BCD$. $ \omega_1$ is a circle situated in $ \triangle ACD$, it is tangent to the segments $ AD$ and $ AC$ at $ M$ and $ N$ respectively, and is also tangent to circle $ \omega$. (i) Show that $ BD\cdot CN+BC\cdot DM=CD\cdot BM$. (ii) Show that $ BM=BC$.
Problem
Source: China Hong Kong Mathematical Olympiad Question 3
Tags: geometry, circumcircle, function, geometry unsolved