27. First, let's assume that $M$ has three vertices on the sides of $\triangle A B C$, or $M$ has one vertex coinciding with a vertex of $\triangle A B C$ (for example, $B$), and another vertex of $M$ is on the opposite side of $B$ (as shown in the figure).

Let the initial state be $\angle A C_{1} B_{1}=\theta_{0}$. We rotate $M$ clockwise and counterclockwise around $C_{1}$. Let $\delta_{1}$ be the angle when $M$ first has a side parallel to a side of $\triangle A B C$ when rotated clockwise, and $\delta_{2}$ be the angle when $M$ first has a side parallel to a side of $\triangle A B C$ when rotated counterclockwise. For $\theta \in\left[\theta_{1}, \theta_{2}\right], \theta_{1}=\theta_{0}-\delta_{1}, \theta_{2}=\theta_{0}+\delta_{2}$, let $M$ first rotate to the corresponding $\theta$ angle around $C_{1}$, and then perform homothetic transformations centered at $A$ and $B$ respectively, so that the corresponding two vertices of the image of $M$ (denoted as $M_{\theta}$) are repositioned on $A C$ and $B C$ respectively. Let $C_{1} B_{1}=m f(\theta), A_{1} B_{1}=n f(\theta), f\left(\theta_{0}\right)=1$, where $m$ and $n$ are the initial distances. Let $\varphi=\angle B+\angle C_{1} B_{1} A_{1}$ (a constant), then $A C=A B_{1}+B_{1} C=\frac{m f(\theta) \sin \theta}{\sin \angle A}+\frac{n f(\theta)}{\sin \angle C} \sin (\varphi-\theta)$. Therefore, $f(\theta)=$ $\frac{A C \sin \angle A \cdot \sin \angle C}{m \sin \theta \cdot \sin \angle C+n \sin (\varphi-\theta) \cdot \sin \angle A}=\frac{A C \sin \angle A \cdot \sin \angle C}{a \sin \left(\theta+\varphi_{1}\right)}$, where $a$ and $\varphi_{1}$ are constants. Since $\sin \left(\theta+\varphi_{1}\right)$ is a convex function, it must achieve its minimum value at the endpoints. Hence, $\max \left\{f\left(\theta_{1}\right), f\left(\theta_{2}\right)\right\} \geqslant f\left(\theta_{0}\right)=1$, so the similarity ratio of $M_{\theta_{1}}$ or $M_{\theta_{2}}$ to $M$ is not less than 1, and they are located within $\triangle A B C$, thus proving the conclusion. For the second case, a similar discussion can be made. Let $B B_{2}=m f(\theta), f\left(\theta_{0}\right)=1$, then $A B_{2}=\frac{B B_{2}}{\sin \angle A} \cdot \sin \theta, C B_{2}=\frac{B B_{2}}{\sin \angle C} \cdot \sin (\angle B-\theta)$. Therefore, $A C=\frac{m f(\theta) \sin \theta}{\sin \angle A}+\frac{m f(\theta)}{\sin \angle C} \sin (\angle B-\theta)=\frac{f(\theta) a \sin \left(\theta+\varphi_{2}\right)}{\sin \angle A \cdot \sin \angle C}$, where $a$ and $\varphi_{2}$ are constants, hence $f(\theta)=\frac{A C \sin \angle A \cdot \sin \angle C}{a \sin \left(\theta+\varphi_{2}\right)}$. The proof can be derived similarly to the previous case.