75th Romanian Mathematical Olympiad

$BC\perp (AB{B}^{\prime })$ and $A{B}^{\prime }\subset (AB{B}^{\prime })$ yields $A{B}^{\prime }\perp BC$. Since $A{B}^{\prime }\perp A^{\prime }B$ (diagonals of the square $AB{B}^{\prime }{A}^{\prime }$), we get $A{B}^{\prime }\perp (A^{\prime }BC)$. Because $A^{\prime }M\subset (A^{\prime }BC)$, it follows that $A^{\prime }M\perp A{B}^{\prime }$. (1)

Let $E\in (AD)$ be such that $AE=BM$. Then $ABME$ is a rectangle, so $AB\parallel ME$. Because $AB\perp (AD{A}^{\prime })$ and $AN\subset (AD{A}^{\prime })$, it follows that $AN\perp ME$. (2)

From $△A^{\prime }AE\equiv △ADN$ (L.L.) one gets $\angle DAN=\angle A{A}^{\prime }E$, whence $\angle A{A}^{\prime }E+\angle A^{\prime }AN=\angle DAN+\angle A^{\prime }AN=90^{\circ }$. Thus, $AN\perp A^{\prime }E$. Using now relation (2) we get $AN\perp (A^{\prime }EM)$, so $AN\perp A^{\prime }M$. Taking into account this last relation and (1), we conclude that $A^{\prime }M\perp (A{B}^{\prime }N)$.