ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.

To prove that PQRS is a rectangle, we need to show that all its angles are 90 degrees.

Step 1: In a rhombus, opposite sides are parallel. So, AB is parallel to DC and AD is parallel to BC.

Step 2: Since P and Q are mid-points of AB and BC respectively, by the Mid-point Theorem, PQ is parallel to AC. Similarly, SR is also parallel to AC.

Step 3: Similarly, since R and S are mid-points of CD and DA respectively, by the Mid-point Theorem, RS is parallel to BD. Similarly, PQ is also parallel to BD.

Step 4: From steps 2 and 3, we can say that PQ is parallel to SR and PS is parallel to QR.

Step 5: Now, we know that in a rhombus, diagonals are perpendicular bisectors of each other. So, angle BAC = 90 degrees and angle ABD = 90 degrees.

Step 6: Since PQ is parallel to AC and AC is perpendicular to BD, by corresponding angles, angle PQS = 90 degrees.

Step 7: Similarly, since PS is parallel to BD and BD is perpendicular to AC, by corresponding angles, angle PSQ = 90 degrees.

Step 8: Therefore, all angles of quadrilateral PQRS are 90 degrees. Hence, PQRS is a rectangle.