In the rectangular parallelepiped $ABCDA'B'C'D'$ we denote by $M$ the center of the face $ABB'A'$. We denote by $M_1$ and $M_2$ the projections of $M$ on the lines $B'C$ and $AD'$ respectively. Prove that: a) $MM_1 = MM_2$ b) if $(MM_1M_2) \cap (ABC) = d$, then $d \parallel AD$; c) $\angle (MM_1M_2), (A B C)= 45^ o \Leftrightarrow \frac{BC}{AB}=\frac{BB'}{BC}+\frac{BC}{BB'}$.