A point $M$ is chosen inside the square $ABCD$ in such a way that $\angle MAC = \angle MCD = x$ . Find $\angle ABM$.
1. Consider the square \(ABCD\) with point \(M\) inside it such that \(\angle MAC = \angle MCD = x\).
2. Let \(L\) be the intersection point of diagonal \(BD\) and line \(MC\).
3. Since \(\angle MAC = \angle MCD = x\), we can use angle chasing to find other angles in the configuration.
4. Note that \(\angle ACD = 90^\circ\) because \(ABCD\) is a square.
5. Since \(\angle MCD = x\), we have \(\angle DCM = 90^\circ - x\).
6. In triangle \(DLC\), \(\angle DLC = 180^\circ - \angle DCL - \angle LDC = 180^\circ - (90^\circ - x) - 90^\circ = 135^\circ - x\).
7. Since \(L\) lies on \(BD\), \(\angle MLB = 135^\circ - x\).
8. In quadrilateral \(AMLB\), \(\angle MLB + \angle MAB = 180^\circ\) because opposite angles in a cyclic quadrilateral sum to \(180^\circ\).
9. Therefore, quadrilateral \(AMLB\) is cyclic.
10. Since \(\angle AML = 135^\circ\) and \(\angle MAB = 45^\circ\), we can conclude that \(M\) is the incenter of triangle \(ACD\).
11. Hence, \(\angle CDM = \angle CDL\).
12. Since \(M = L\) lies on diagonal \(BD\), we have \(\angle ABM = 45^\circ\).

The final answer is \(\boxed{45^\circ}\).