If a line intersects two concentric circles with centre A in points P, Q, R and S respectively, then prove that PQ=RS.

To prove that \( PQ = RS \), consider the following setup:

- Let the two concentric circles have the same center \( A \).
- A line intersects the smaller circle at points \( P \) and \( Q \), and the larger circle at points \( R \) and \( S \).

### Step-by-Step Proof:
1. **Draw the radii:**
Draw radii \( AP, AQ, AR, \) and \( AS \) from the center \( A \) to points \( P, Q, R, \) and \( S \), respectively. Since \( AP = AQ \) (radii of the smaller circle) and \( AR = AS \) (radii of the larger circle), we have the relationships:
\[
AP = AQ \quad \text{and} \quad AR = AS.
\]

2. **Perpendicular bisector:**
The line passing through \( P, Q, R, S \) intersects the center \( A \) perpendicularly. Thus, \( A \) is equidistant from both the midpoints of segments \( PQ \) and \( RS \).

3. **Similar triangles:**
The triangles formed by the radii \( AP, AQ, AR, \) and \( AS \) and the segments \( PQ \) and \( RS \) are congruent because:
- Both share the same center \( A \).
- The angles subtended by the intersecting line are equal.
- The radii are equal.

4. **Equality of chords:**
The distances \( PQ \) and \( RS \) are measured on the same line and subtend equal angles at the center \( A \). Therefore, \( PQ = RS \).

Thus, the lengths of the chords \( PQ \) and \( RS \) are equal, as required.