2. In the plane, a square $A B C D$ is given. Circle $k$ passes through points $A, B$ and is tangent to line $C D$. Let $M(M \neq B)$ be the intersection of circle $k$ and side $B C$. Determine the ratio $|C M|$ : $:|B M|$.
2. Since the center of circle $k$ lies on the axis of side $A B$, which is also the axis of the opposite side $C D$, circle $k$ touches segment $C D$ at its midpoint $S$ (Fig. 1). Since angle $A B M$ is a right angle, $A M$ is the diameter of circle $k$, and therefore angle $A S M$ is also a right angle. It follows that $|\Varangle D S A|=90^{\circ}-|\Varangle C S M|=|\Varangle S M C|$, so triangles $S M C$ and $A S D$ are similar, which means $|C M|:|C S|=|D S|:|D A|$. If we let $a=|D A|$ and $x=|C M|$, then $x: \frac{1}{2} a=\frac{1}{2} a: a$, thus $x=\frac{1}{4} a$. Therefore, $|C M|:|B M|=1: 3$.

Let us add that the equality $x=\frac{1}{4} a$ can also be derived from the Pythagorean theorem for triangles $A M B, A M S$:

$$
|A B|^{2}+|B M|^{2}=|A M|^{2}=|A S|^{2}+|S M|^{2},
$$

so

![](https://cdn.mathpix.com/cropped/2024_04_17_c904719f0fd164a408cdg-2.jpg?height=504&width=507&top_left_y=800&top_left_x=1362)

Fig. 1

$$
a^{2}+(a-x)^{2}=\left(a^{2}+\left(\frac{1}{2} a\right)^{2}\right)+\left(\left(\frac{1}{2} a\right)^{2}+x^{2}\right)
$$

from which, after simplification,

$$
x=\frac{1}{4} a \text {. }
$$

Note. If the student is familiar with the concept of the power of a point with respect to a circle, they can write $|C M| \cdot |C B|=|C S|^{2}$, from which it immediately follows that $|C M|=\frac{1}{4} a$.