1. Translate $B M$ to $A N$, connect $M N, D N$, then so $\triangle A D N \cong \triangle B M C$
So $\angle A N D=\angle B M C, \angle A D N=\angle B C M$
Since $\angle B M C+\angle A M D=180^{\circ}$
So $\angle A N D+\angle A M D=180^{\circ}$
So $A, N, D, M$ are concyclic.
So $\angle A M N=\angle A D N=\angle B C M$
Since $M N / / B A$
So $\angle B A M=\angle A M N=\angle B C M$
Since $\angle B A M+\angle M A D=\angle B A D=90^{\circ}$
So $\angle B C M+\angle D A M=90^{\circ}$
1. In the interior of rectangle $A B C D$, take a point $M$ such that $\angle B M C + \angle A M D = 180^{\circ}$. Try to find the size of $\angle B C M + \angle D A M$.