$ \omega $ is the circumcircle of triangle ???

From where did you get this problem? I could not find it at the Hong Kong MO webpage http://www.edb.gov.hk/index.aspx?nodeid=6351

Please check the problem statement, I wonder if you mean $\omega_1$ is inside the "angle" instead of "triangle"

Let the tangent point of $\omega_{1}$ with center $S$, radius $r$ and the circumcircle of $CBD$ with center $E$ be $T$. it is obvious that $S, T, E$ as well as $N, T, B$ are collinear. Let the sides of $\Delta ABC$ be $a, b, c$. As $tanA = a/b$, it is easy to get $tan(A/2) = (c-b)/a$. $DB = \frac{a^{2}}{c}; AD = \frac{b^{2}}{c}; AM = AN = \frac{ra}{c-b}; CN = \frac{-b^{2}+bc-ra}{c-b}$; and of course, $SE = r+a/2$. Now consider the right-angled trapezium $NSEC$, we have: $CN^{2} = (a/2+r)^{2} - (a/2-r)^{2}$, or $(\frac{-b^{2}+bc-ra}{c-b})^{2} = 2ra$. The equation gives us two possible values of $r$: $r_{1} = \frac{(c-b)(c-a)}{a}, r_{2} =\frac{(c-b)(c+a)}{a}$. This fact tells us there are two such circles, the smaller one, $\omega_{1}$, is tangent to sides $AC, AB$ and the circumcircle of $CDB$ within them, the second one is tangent to $AC$ and $AB$ prolonged and to the circumcircle of $CDB$ on the opposite side. Hence $AM = AN = c-a; CN = b+a-c; MD =\frac{a(c-a)}{c}$. With all the needed lenghts found, there is no problem to prove the two equations in the questions. Note: Another way to find all the needed lengths without computing $r$ is to prove that the tangent point of the incircle of $ABC$ and side $AC$ is the midpoint of $CN$.

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As a remark: proving first the second point, will speed-up the calculations. And this can be done fast by inversion: of pole $B$ and power $BC^2$ sends the circle $\odot(BCD)$ to the line $AC$ and preserves the circle $\odot (MNT)$, hence $BM=BC$ and $(2)$ has been proved. Consequently $CN=b+c-a$ and indeed, the incircle of $\Delta ABC$ touches $AC$ at the midpoint of $CN$, a.s.o. Best regards, sunken rock

Hello sunken rock, I only want to say that once we got $CN = b+c-a$, there is no need to show that the incircle of $ABC$ touches side $AC$ at the midpoint of $CN$. All the needed lenghts are ready to use. In my note I meant to prove this property is a way to get all the lenghts.

sorry to revive.But I solved the second part don't know if i am snipped or not.For the first part I thought it can be done by casey's theorem.but could not solve it.I used the notation of vslmat. here goes, Denote $T$ be the common point of $\omega$ and $\omega_{1}$ and $S,T,E$ are collinear. $\angle TNC + \angle TCN = \frac{1}{2} (\angle TSN +\angle TEC)=\frac{1}{2}(360^{\circ} -\angle SNC -\angle ECN) =90^{\circ}$ $ \implies \angle NTC =90^{\circ}$ From this we conclude that $B,T,N$ are co-linear.From power of point we get, $BM^{2}=BT.BN$ and it is given that $\angle C =90^{\circ}$ and $CT \perp TN$.so $BC^2=BT.BN \implies BM=BC$

The first one follows from Casey,s theorem to the circles $B,C,\omega_1,D$ with respect to $\omega$. The second one follows from some bashing.......(I mean I did it...)