In the square $A B C D$, the point $M$ is located such that $D C M \angle=M A C \angle=25^{\circ}$. What is the measure of the angle $A B M$?
I. Solution. Let the side of the square be of unit length.

![](https://cdn.mathpix.com/cropped/2024_05_02_590929d383f8cc692e22g-1.jpg?height=517&width=466&top_left_y=224&top_left_x=819)

Clearly,

$$
\begin{aligned}
& \angle ACM = \angle ACD - \angle MCD = 45^\circ - 25^\circ = 20^\circ, \\
& \angle BAM = \angle BAC + \angle CAM = 45^\circ + 25^\circ = 70^\circ, \\
& \angle AMC = 180^\circ - \angle ACM - \angle CAM = 135^\circ.
\end{aligned}
$$

By the Law of Sines in triangle \(ACM\):

$$
AM = \frac{\sin 20^\circ}{\sin 135^\circ} \cdot AC = \frac{\sin 20^\circ}{\frac{1}{\sqrt{2}}} \cdot \sqrt{2} = 2 \sin 20^\circ
$$

Applying the Law of Cosines to side \(BM\) in triangle \(ABM\):

$$
BM^2 = AM^2 + AB^2 - 2 \cdot AB \cdot AM \cdot \cos 70^\circ = 4 \sin^2 20^\circ + 1 - 4 \sin 20^\circ \cos 70^\circ = 1
$$

Thus, \(BM = 1\), and therefore triangle \(ABM\) is isosceles: \(AB = BM\), which implies \(\angle BMA = \angle BAM = 70^\circ\). Consequently, \(\angle ABM = 40^\circ\).

II. Solution. By redrawing the diagram, it can be conjectured that the circumcenter of triangle \(AMC\) is point \(B\).

![](https://cdn.mathpix.com/cropped/2024_05_02_590929d383f8cc692e22g-1.jpg?height=518&width=463&top_left_y=1649&top_left_x=820)

First, we verify this. Let the circumcircle of triangle \(AMC\) be \(k\), and its center be \(O\). In circle \(k\), the inscribed angle corresponding to arc \(MC\) is \(\angle MAC = 25^\circ\). Since \(\angle MCD = 25^\circ\), this angle is the inscribed angle of the tangent to arc \(MC\), and thus \(CD\) is the tangent to circle \(k\) at point \(C\). Therefore, \(CD \perp CO\), which means \(O\) lies on line \(BC\).

Moreover, \(O\) lies on the perpendicular bisector of segment \(CA\), which is line \(BD\).

Thus, \(O\) is the intersection of lines \(BC\) and \(BD\), which is point \(B\).

This implies that \(BA = BM\), so triangle \(BAM\) is isosceles. We know that \(\angle BAM = 25^\circ + 45^\circ = 70^\circ\), hence \(\angle ABM = 180^\circ - 2 \cdot 70^\circ = 40^\circ\).