Find the angle between the lines $A C$ and $B D$, if the distance between the midpoints of segments $A D$ and $B C$ is equal to the distance between the midpoints of segments $A B$ and $C D$.

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Let $M, N, K$ and $L$ be the midpoints of segments $AD, BC, AB$ and $CD$ respectively. Since $ML$ and $KN$ are the midlines of triangles $ADC$ and $ABC$ with the common base $AC$, then $ML \| KN$. Therefore, points $M, N, K$ and $L$ lie in the same plane. Moreover, $MK \| LN$, so quadrilateral $KNLM$ is a parallelogram. According to the problem, its diagonals $KL$ and $MN$ are equal, hence $KNLM$ is a rectangle. Therefore, the angle between lines $AC$ and $BD$ is equal to the angle between the parallel lines $KN$ and $KM$, i.e., $90^{\circ}$.

## Answer

$90^{\circ}$