Four, (15 points) As shown in the figure, in $\triangle A B C$, $\angle B=90^{\circ}, M$ is a point on $A B$ such that $A M = B C, N$ is a point on $B C$ such that $C N = B M$. Connect $A N, C M$ intersecting at point $P$. Try to find the degree of $\angle A P M$ and write out your reasoning and proof process.
IV. $\angle A l^{\prime} M=45^{\circ}$. Hint:
Solution 1 Draw a perpendicular to $BC$ through $C$, and intercept $CD=AM=BC$ on it, then connect $DN, DA, DM$. It can be proven that $\triangle DCN \cong \triangle CBM$. Thus, $\triangle ADN$ is an isosceles right triangle.

Solution 2 Outside $\triangle ABC$, draw a perpendicular to $AB$ through $M$, and intercept $ME=MB=CN$ on it, then connect $AE, NE, MN$. It is easy to prove that $\triangle CMN \cong \triangle EMN$, hence $CM \parallel EN$. Through calculation, it can still be deduced that $\triangle AEN$ is an isosceles right triangle.

Solution 3 Draw a perpendicular to $AB$ through $A$, and intercept $AK=CN=MB$ on it, then connect $KM, KC$. It is easy to know that $\triangle KMC$ is an isosceles right triangle. Also, $\triangle AKC \cong \triangle CNA$, hence $KC \parallel AN$.