Diagonals of trapezium  $ABCD$ are mutually perpendicular and the midline of the trapezium is $5$. Find the length of the segment that connects the midpoints of the bases of the trapezium.
1. Let $ABCD$ be a trapezium with $AB \parallel CD$ and diagonals $AC$ and $BD$ intersecting at point $O$ such that $AC \perp BD$.
2. Let $E$ and $G$ be the midpoints of $AB$ and $CD$ respectively. The line segment $EG$ is the midline of the trapezium, and it is given that $EG = 5$.
3. Let $F$ and $H$ be the midpoints of the non-parallel sides $AD$ and $BC$ respectively.
4. According to the Midpoint Theorem for quadrilaterals, the quadrilateral $EFGH$ formed by connecting the midpoints of the sides of any quadrilateral $ABCD$ is a parallelogram.
5. Since the diagonals $AC$ and $BD$ of trapezium $ABCD$ are perpendicular, the parallelogram $EFGH$ becomes a rectangle.
6. In a rectangle, the diagonals are equal in length. Therefore, the length of the segment $FH$ (which connects the midpoints of the non-parallel sides $AD$ and $BC$) is equal to the length of the midline $EG$.
7. Given that $EG = 5$, it follows that $FH = 5$.

Therefore, the length of the segment that connects the midpoints of the bases of the trapezium is $\boxed{5}$.