Solution:

Note that triangle $M C A$ is a right isosceles triangle with $\angle A C M=90^{\circ}$ and $\angle M A C=45^{\circ}$. Since $\angle A B C=90^{\circ}$, there is a circle $k$ with diameter $A C$ which also passes through points $B$ and $C$. As inscribed angles, $\angle M A C=\angle A B C$, thus the measure of $\angle M B C=45^{\circ}$.

Solution 2: Place 3 copies of triangle $A B C$ on the square as shown below.

Clearly the new diagram is a large square (which can be proven easily by looking at the angles of the copies of the triangle). The diagonals of this large square meet at $M$. By symmetry, $\angle M B C=45^{\circ}$.

Solution 3: Place triangle $A B C$ on coordinate axes so that $A=(0,2 c), B=(0,0), C=(2 a, 0)$. Let $N$ be the midpoint of hypotenuse $A C$ and draw line $\ell$ through $N$ parallel to $B C$ which meets $A B$ at $F$. Drop a perpendicular from $M$ to $B C$ which meets $\ell$ at $G$. It is evident that triangles $A F N$ and $N G M$ are congruent.
Thus $N G=A F=c$ and $M G=F N=a$. Consequently, the coordinates of $M$ are $(a, c)+(c, a)=(a+c, a+c)$, so the line from $M$ to the origin (at $B$) has a slope of 1.