Estonian Mathematical Olympiad

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

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Alternative solution.

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