Let the three squares touch at point O. Let ABCD be the fourth square, and A be the common vertex of the three squares. Let c be the circle with center O and radius OA. We have ∠BCA=45°, and then C∈c. Therefore, ∠ACM=90°, where AM is the diameter of c. Since ∠ACD=45°, we get ∠DCM=45°.