Let $\angle DAL = \angle LAC = \alpha$. Then $\angle FAE = 90^\circ - 2\alpha$ (Fig. 1).

As $AE = AF$, we obtain $\angle AFE = \frac{180^\circ - (90^\circ - 2\alpha)}{2} = 45^\circ + \alpha$.

But as $M$ bisects the hypotenuse $AL$ of the right triangle $ALD$, it follows that $M$ is the circumcentre of $ALD$, implying $MA = MD$. Hence $\angle MDA = \angle MAD = \alpha$, implying $\angle AFE = \angle AFD = 90^\circ - \alpha$.

We obtain $45^\circ + \alpha = 90^\circ - \alpha$ which implies $\angle CAD = 2\alpha = 45^\circ$. Hence also $\angle ACD = 45^\circ$, implying $AD = CD$. Consequently, $ABCD$ is a square.

Fig. 1

As $AE = AF$, the triangle $AEF$ is isosceles with vertex angle at $A$ (Fig. 1). Denote $\angle AEF = \angle AFE = \beta$.

Note that triangles $MAF$ and $MLD$ are equal because $\angle FMA = \angle DML$, $\angle MAF = \angle MLD$ and $MA = ML$. Thus $AF = DL$ which shows that $ADLF$ is a rectangle. As the diagonals of a rectangle are equal and bisect each other, we have $MA = MF$, implying that the triangle $MAF$ is isosceles with vertex angle at $M$. Hence $\angle MAF = \angle MFA = \beta$.

As $\angle EAF = 180^\circ - 2\beta$, we have $\angle MAE = \beta - (180^\circ - 2\beta) = 3\beta - 180^\circ$. Hence also $\angle MAD = 3\beta - 180^\circ$, because $AM$ bisects the angle $DAE$. We obtain the equation $\beta + (3\beta - 180^\circ) = 90^\circ$ which gives $\beta = 67.5^\circ$. Thus $\angle BAC = 180^\circ - 2 \cdot 67.5^\circ = 45^\circ$, implying that the diagonal of the rectangle $ABCD$ bisects its angle. Consequently, $ABCD$ is a square.

Fig. 1