3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q are the mid-points of AB and BC respectively , PQ || AC and PQ = ½ AC (Midpoint theorem) — (i) In ΔADC, SR || AC and SR = ½ AC (Midpoint theorem) — (ii) So, PQ || SR and PQ = SR As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram. , PS || QR and PS = QR (Opposite sides of parallelogram) — (iii) Now, In ΔBCD, Q and R are mid points of side BC and CD respectively. , QR || BD and QR = ½ BD (Midpoint theorem) — (iv) AC = BD (Diagonals of a rectangle are equal) — (v) From equations (i), (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved
