4. In the rectangle $A B C D$, point $M$ lies on side $C D$ and it holds that $D M=2 \cdot C M$, and $O$ is the intersection of the diagonals. The lines $A C$ and $B M$ intersect at a right angle. Determine $\measuredangle B O M$.
Solution. Let $E$ be the midpoint of side $B C$. Then $O E \perp B C$. According to this, point $H$ is the orthocenter of triangle $O B C$, so $C H \perp O B$. Meanwhile, $O H$ is the midline of triangle $D B M$, so $O H=\frac{1}{2} D M=M C$. Therefore, quadrilateral

![](https://cdn.mathpix.com/cropped/2024_06_05_ac2bd3b4ef21cfd30547g-3.jpg?height=279&width=407&top_left_y=1802&top_left_x=1075)
$A$

$O H C M$ is a parallelogram, hence it follows that $O M \perp B O$.