30. (Su Kuan 1)

Proof: Let $\triangle_{1}$ be the area of $\triangle A B M$, $s_{1}$, the semi-perimeter of $\triangle A B M$, and $r_{1}$ the radius of the incircle of $\triangle A B M$: $\triangle_{2}$ is the area of $\triangle B M C$, $s_{2}$ is the semi-perimeter of $\triangle B M C$, and $r_{2}$ is the radius of the incircle of $\triangle B M C$. For $\triangle A B C$, $\triangle, s, r$ represent the corresponding values. Let $P^{\prime}$ be the point where the incircle of $\triangle A B M$ touches side $A B$, $Q^{\prime}$ the point where the incircle of $\triangle B M C$ touches side $B C$; $P, Q$ are the points where the incircle of $\triangle A B C$ touches sides $A B, B C$ respectively.
Then $\triangle_{1}=s_{1} r_{1}, \triangle_{2}=s_{2} r_{2}, \triangle=s r$.
$$\text { Thus, } s r=s_{1} r_{1}+s_{2} r_{2}=r^{\prime}\left(s_{1}+{ }_{2} s\right) \text {, }$$

where $\gamma^{\prime}=\gamma_{1}=\gamma_{2}$.
On the other hand, by the similarity of the triangles, we get
$$\frac{A P^{\prime}}{A P}=\frac{r^{\prime}}{r}, \frac{C Q^{\prime}}{C Q}=\frac{r^{\prime}}{r} \text {. }$$

Moreover, $A P=s-B C, C Q=s-A B, A P^{\prime}$
$$=s_{1}-B M, C Q^{\prime}=s_{2}-B M \text {. }$$

From (1) and (2), $\frac{r^{\prime}}{r}=\frac{s_{1}-B M}{s-B C}$.
$$=\frac{s_{2}-B M}{s-A B} . \\ \text { Also, } \frac{r^{\prime}}{r}=\frac{s_{1}+s-2 B M}{2 s-A B-B C}=\frac{s-B M}{A C}, \\ \frac{r^{\prime}}{r}=\frac{s-B M}{A C}=\frac{s}{+B M}, \\ s^{2}-B M^{2}=s \cdot A C \text { or } B M^{2}=s(s-$$
$A C$ ).
But $\mathrm{s}-A C=B P, \frac{B P}{r}=\operatorname{ctg} \frac{B}{2}$,
so $B P=r \operatorname{ctg} \frac{B}{2}$.
Thus $B M^{2}=s(s-A C)=s \cdot B P$
$$=s \cdot r \cdot \operatorname{ctg} \frac{B}{2}=\Delta \operatorname{ctg} \frac{B}{2} .$$