Note that the shorter the distance from the center $O$ of the circle to a chord, the longer the length of the chord. Since there are a finite number of chords, there is a shortest one, say $A B$. By the condition, it passes through the midpoint $K$ of some other chord, say $C D$. If the intersection point of $A B$ and $C D$ is not also the midpoint of $A B$, then the distance from point $O$ to $C D$ will obviously be greater than the distance from $O$ to $A B$ (since $OK$ will be longer than the length of the perpendicular dropped from point $O$ to $A B$), hence, the chord $C D$ has a smaller length than $A B$, which is a contradiction. Therefore, $C D$ passes through the midpoint of $A B$, from which the perpendiculars dropped from point $O$ to these chords coincide. This is possible only if $A B$ and $C D$ coincide, or if $A B$ and $C D$ intersect at the center. The first is impossible, so $A B$ and $C D$ are diameters. The same can be proven for all other chords passing through $O$.

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Fig. 1