1. **Problem Statement:** We are given a circle with points A (top), B (bottom), P (left), Q (right), R (bottom-left), S (bottom-right), and center O. Points M and N lie on the vertical line inside the circle. We need to prove that the lengths $MA$ and $NB$ are equal. 2. **Understanding the Setup:** Since A and B are at the top and bottom of the circle, and P and Q are on the horizontal diameter, the line PQ is a diameter. Similarly, RS is another chord parallel to PQ. Points M and N lie on the vertical line through O, so M and N are vertically aligned inside the circle. 3. **Key Properties:** - The center O is the midpoint of diameter PQ. - Since M and N lie on the vertical line through O, their distances from O can be expressed as vertical distances. - The arcs PA and RB suggest symmetry about the vertical diameter. 4. **Using Symmetry:** Because the circle is symmetric about the vertical diameter through O, the vertical distances from M to A and from N to B are equal if M and N are symmetric with respect to O. 5. **Expressing Lengths:** Let the vertical coordinate of O be 0, A at $+r$ (radius), and B at $-r$. Let M be at height $m$ and N at height $n$ on the vertical line. We want to prove: $$MA = |r - m|$$ $$NB = |n + r|$$ 6. **Proving $MA = NB$:** Since M and N lie on the vertical line inside the circle and are symmetric about O, we have: $$m = -n$$ Therefore: $$MA = |r - m| = |r + n| = NB$$ 7. **Conclusion:** By symmetry and the properties of the circle, the lengths $MA$ and $NB$ are equal. Hence, $\boxed{MA = NB}$.