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

Let us draw the figure with given condition as
In ΔABC, P and Q are the mid- points of AB and BC. PQ || AC and by using mid - point theorem PQ=1/2AC (1)
Similarly, in ΔADC, R and S are the mid- points of CD and AD. SR || AC and by using mid point theorem SR=1/2AC (2)
From (1) and (2), we get PQ || RS and PQ=SR
Now, in quadrilateral PQRS its one pair of opposite sides PQ and SR is equal and parallel.
Therefore, PQRS is a parallelogram
AB=BC (Sides of a rhombus) ⇒ 1/2AB=1/2BC
PB=BQ ∠3=∠4
Now, in ΔAPS and ΔCQR, we have
AP=CQ (Halves of equal sides AB, BC)
AS=CR (Halves of equal sides AD, CD)
PS=QR (Opp. sides of parallelogram PQRS)
Therefore, ΔAPS≅ΔCQR using SSS Congruency Theorem
∠1=∠2 (Corresponding parts of congruent triangles are equal)
Now , ∠1+∠SPQ+∠3=180° (Linear pair axiom)
Therefore, ∠1+∠SPQ+∠3=∠2+∠PQR+∠4
But, ∠1=∠2 and ∠3=∠4
Therefore, ∠SPQ=∠PQR (3)
Since, SP || RQ and PQ intersects them
Therefore, ∠SPQ+∠PQR=180° (4) (Since consecutive interior angles are supplementary)
From (3) and (4), we get
∠PQR+∠PQR=180°
2∠PQR=180°
∠PQR=90°
∠SPQ=∠PQR=90°
Thus, PQRS is a parallelogram whose one angle ∠SPQ=90°.
Hence PQRS is a rectangle.