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.
In ΔABC, P and Q are the mid-points of sides AB and BC respectively.
∴ PQ || AC and PQ = AC (Using mid-point theorem) … (1)
In ΔADC,
R and S are the mid-points of CD and AD respectively.
∴ RS || AC and RS = AC (Using mid-point theorem) … (2)
From equations (1) and (2), we obtain
PQ || RS and PQ = RS
Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to
each other, it is a parallelogram.
Let the diagonals of rhombus ABCD intersect each other at point O.
In quadrilateral OMQN,
MQ || ON ( 
PQ || AC)
QN || OM ( 
QR || BD)
Therefore, OMQN is a parallelogram.
⇒ ∠MQN = ∠NOM
⇒ ∠PQR = ∠NOM
However, ∠NOM = 90° (Diagonals of a rhombus are perpendicular to each other)
∴ ∠PQR = 90°
Clearly, PQRS is a parallelogram having one of its interior angles as 90º.
Hence, PQRS is a rectangle.
