Given a parallelogram $A B C D$, where $A B=a, A D=b$. The first circle has its center at vertex $A$ and passes through $D$, the second circle has its center at $C$ and passes through $D$. An arbitrary circle with center $B$ intersects the first circle at points $M_{1}, N_{1}$, and the second circle at points $M_{2}, N_{2}$. What is the ratio $M_{1} N_{1}: M_{2} N_{2}$?
Points $M_{1}, N_{1}$ are symmetric with respect to the line $A B$, so $M_{1} N_{1}$ is equal to twice the distance from $M_{1}$ to $A B$. Similarly, $M_{2} N_{2}$ is equal to twice the distance from $M_{2}$ to $B C$. Moreover, $C M_{2}=C D=A B, A M_{1}=A D=B C$, $B M_{1}=B M_{2}$, and thus, triangles $A B M_{1}$ and $C M_{2} B$ are equal. Therefore, the desired ratio, which is the ratio of the heights of these triangles, is the inverse of the ratio of the corresponding sides, that is, $b: a$.

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